Contents / Abstract Algebra / Subgroups, Cosets, and Quotient Groups

Chapter 2

Subgroups, Cosets, and Quotient Groups

The one-step and two-step subgroup tests; cosets proved to partition a group into equal blocks, and Lagrange's theorem proved from that partition with Euler and Fermat as corollaries; four equivalent characterisations of normality; quotient groups proved well defined exactly when the subgroup is normal; group actions, orbit-stabiliser and the class equation; and the Sylow theorems with applications to groups of small order.

Introduction

A group, as built in Groups, is a set with one associative operation, an identity and inverses. That is a small amount of data, and on its own it says almost nothing about how large or complicated the object is. Everything quantitative in finite group theory comes from a single further idea: a subgroup $H \leq G$ cuts $G$ into translated copies of itself, all of the same size, with no overlap. Counting those copies converts an algebraic containment into an arithmetic divisibility, and that conversion — Lagrange's theorem — is the engine of this chapter and of most of the classification results that follow it.

The chapter has four movements. The first sets up subgroups and the tests that let you certify one in a single line instead of re-checking four axioms; it also explains why intersections of subgroups are subgroups while unions are almost never, which is the fact that makes "the subgroup generated by a set" well defined. The second builds cosets, proves that they partition the group into equal-sized blocks, and extracts Lagrange's theorem together with the corollaries that make it useful: the order of every element divides the order of the group, every group of prime order is cyclic, and — reading the whole thing inside $(Z / n)^{\times}$ — Euler's and Fermat's theorems drop out as special cases of a statement about symmetry.

The third movement asks when the set of cosets can itself be made into a group. The answer is: exactly when the subgroup is normal, and the reason is a single well-definedness check on the proposed multiplication $(a N) (b N) = ab N$ . That check is the crux of the chapter. Multiplication of cosets is defined by choosing representatives, and a definition that depends on the choice is not a definition at all; normality is precisely the condition that the choice does not matter. Everything else about quotients — the correspondence with kernels, the isomorphism theorems in Group Homomorphisms and Isomorphisms, the construction of $Z / n Z$ — is downstream of that one verification.

The fourth movement is the partial converse to Lagrange. Lagrange says $∣ H ∣$ divides $∣ G ∣$ ; it does not say that every divisor of $∣ G ∣$ is realised by a subgroup, and $A_{4}$ , of order $12$ with no subgroup of order $6$ , is the standing counterexample. The Sylow theorems repair as much of the converse as is true: for each prime power $p^{a}$ exactly dividing $∣ G ∣$ , subgroups of that order exist, are all conjugate, and their number is pinned down by two congruence-and-divisibility conditions that are usually strong enough to determine it outright. That last point is what makes Sylow a computational tool rather than an abstract existence statement, and the closing section uses it to prove that no group of order $pq$ is simple.

Throughout, $G$ is a group written multiplicatively with identity $e$ , and $∣ G ∣$ denotes its order, finite unless stated otherwise.

2.1Subgroups and the subgroup tests

A subset of a group inherits the operation but not, in general, the axioms: the even integers are closed under addition, the odd integers are not; the invertible $2 \times 2$ matrices form a group, the matrices of determinant $2$ do not. The subsets that survive are the ones worth naming.

Definition 2.1 (Subgroup). A subset $H \subseteq G$ is a subgroup of $G$ , written $H \leq G$ , if $H$ is a group under the operation of $G$ restricted to $H$ . Explicitly: $e \in H$ ; $ab \in H$ whenever $a, b \in H$ ; and $a^{- 1} \in H$ whenever $a \in H$ .

$H$ is a proper subgroup if $H \neq = G$ , and nontrivial if $H \neq = {e}$ .

Associativity is not listed because it is inherited free of charge: an identity that holds for all elements of $G$ holds in particular for all elements of $H$ . This is why the check is three conditions rather than four, and it is also why "sub-object" definitions in algebra are always shorter than the parent definition — universally quantified laws restrict automatically, while existence assertions such as "there is an identity" do not.

Two further remarks fix the identity question. If $H \leq G$ then the identity of $H$ is the identity of $G$ : if $f \in H$ satisfies $ff = f$ , then multiplying by $f^{- 1}$ in $G$ gives $f = e$ . So a subgroup cannot quietly carry its own private identity, and likewise inverses computed in $H$ agree with inverses computed in $G$ .

Proposition 2.2 (Two-step subgroup test). A subset $H \subseteq G$ is a subgroup if and only if

$H \neq = \emptyset$ ,
$ab \in H$ for all $a, b \in H$ (closure under the operation), and
$a^{- 1} \in H$ for all $a \in H$ (closure under inversion).

Proof. If $H \leq G$ then $e \in H$ , so $H$ is nonempty, and conditions (2) and (3) are part of Definition Subgroup.

Conversely, suppose (1)–(3) hold. By (1) choose $a \in H$ ; by (3) $a^{- 1} \in H$ ; by (2) $a a^{- 1} = e \in H$ . So $H$ contains the identity, is closed under the operation by (2), and contains inverses by (3). Associativity is inherited from $G$ . Hence $H$ is a group under the restricted operation, i.e. $H \leq G$ .∎

The nonemptiness hypothesis is not decoration. The empty set satisfies (2) and (3) vacuously and is not a group, because a group must have an identity. In practice one verifies (1) by exhibiting $e \in H$ , which is usually the easiest of the three checks and simultaneously the one most often forgotten.

The three conditions can be compressed into one, and for finite calculations the compressed form saves real work.

Theorem 2.3 (One-step subgroup test). A subset $H \subseteq G$ is a subgroup if and only if $H \neq = \emptyset$ and

a b^{- 1} \in H for all a, b \in H .

Proof. ( $\Rightarrow$ ) If $H \leq G$ and $a, b \in H$ , then $b^{- 1} \in H$ and hence $a b^{- 1} \in H$ by closure.

( $\Leftarrow$ ) Suppose $H$ is nonempty and closed under $(a, b) \mapsto a b^{- 1}$ . Pick $a \in H$ .

Identity. Taking $b = a$ gives $a a^{- 1} = e \in H$ .

Inverses. For $b \in H$ , take the pair $(e, b)$ : since $e \in H$ and $b \in H$ , we get $e b^{- 1} = b^{- 1} \in H$ .

Closure. For $a, b \in H$ we now know $b^{- 1} \in H$ , so applying the hypothesis to the pair $(a, b^{- 1})$ gives $a (b^{- 1})^{- 1} = ab \in H$ .

All three conditions of the proposition Two-step subgroup test hold, so $H \leq G$ .∎

Note the order of the three steps: each uses the conclusion of the previous one. The hypothesis is applied three times to three different pairs, which is exactly why one condition suffices — $a b^{- 1}$ mixes multiplication and inversion, so specialising one argument to $e$ isolates each operation in turn. In additive notation the test reads $a - b \in H$ , which is the form one uses for subgroups of $Z$ or subspaces of a vector space.

Example 2.4 (The even integers). Show that $2 Z = {2 k : k \in Z}$ is a subgroup of $(Z, +)$ , and that the odd integers are not.

Solution.

$2 Z$ is nonempty: $0 = 2 \cdot 0$ lies in it.
For $a = 2 j$ and $b = 2 k$ , the additive form of the theorem One-step subgroup test asks for $a - b \in 2 Z$ : indeed $2 j - 2 k = 2 (j - k)$ .
Hence $2 Z \leq Z$ .
The odd integers fail at once: $1 - 1 = 0$ is not odd, so the set is not closed under subtraction, and it does not contain the identity $0$ either.

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Example 2.5 (Determinant-one matrices). Let $G = G L_{2} (R)$ under matrix multiplication. Show that $S L_{2} (R) = {A : det A = 1}$ is a subgroup, and that ${A : det A = 2}$ is not.

Solution.

$I \in S L_{2} (R)$ since $det I = 1$ , so the set is nonempty.
If $det A = det B = 1$ then $det (A B^{- 1}) = det A \cdot (det B)^{- 1} = 1 \cdot 1 = 1$ , so $A B^{- 1} \in S L_{2} (R)$ . By the theorem One-step subgroup test it is a subgroup.
For the second set, $det I = 1 \neq = 2$ , so it does not contain the identity and cannot be a subgroup. Equivalently, $det (A B) = 4$ for two such matrices, so it is not even closed.
The pattern generalises: the preimage of a subgroup under a homomorphism is a subgroup, and ${1} \leq R^{\times}$ while ${2}$ is not a subgroup of $R^{\times}$ .

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Proposition 2.6 (Finite subsets need only closure). Let $G$ be a group and $H \subseteq G$ a nonempty finite subset closed under the operation of $G$ . Then $H \leq G$ .

Proof. Let $a \in H$ . The powers $a, a^{2}, a^{3}, \dots$ all lie in $H$ by closure, and $H$ is finite, so two of them coincide: $a^{i} = a^{j}$ with $i < j$ . Cancelling in $G$ gives $a^{j - i} = e$ with $j - i \geq 1$ , so $e = a^{j - i} \in H$ . If $j - i = 1$ then $a = e$ and $a^{- 1} = e \in H$ ; otherwise $a^{- 1} = a^{j - i - 1} \in H$ , again a positive power of $a$ . So $H$ contains $e$ and is closed under inverses, and the proposition Two-step subgroup test applies.∎

Pitfall. Finiteness is essential. The set $N = {1, 2, 3, \dots}$ inside $(Z, +)$ is nonempty and closed under addition, and is emphatically not a subgroup: it has no inverses. The proof above needs the pigeonhole step, which is exactly where infiniteness breaks it.

Definition 2.7 (Centre and centraliser). The centre of $G$ is

Z (G) = {z \in G : z g = g z for all g \in G},

and the centraliser of $x \in G$ is $C_{G} (x) = {g \in G : gx = xg}$ .

Proposition 2.8 (Centres and centralisers are subgroups). For every group $G$ and every $x \in G$ , both $Z (G)$ and $C_{G} (x)$ are subgroups of $G$ , and $Z (G) = ⋂_{x \in G} C_{G} (x)$ .

Proof. We check $C_{G} (x)$ with the theorem One-step subgroup test. It is nonempty since $e x = x = x e$ . Let $a, b \in C_{G} (x)$ . From $b x = x b$ we get $x b^{- 1} = b^{- 1} x$ by multiplying on both sides by $b^{- 1}$ . Hence

(a b^{- 1}) x = a (b^{- 1} x) = a (x b^{- 1}) = (a x) b^{- 1} = (x a) b^{- 1} = x (a b^{- 1}),

so $a b^{- 1} \in C_{G} (x)$ and $C_{G} (x) \leq G$ .

An element $z$ commutes with every $g \in G$ precisely when $z \in C_{G} (g)$ for every $g$ , which is the asserted intersection formula. That $Z (G)$ is a subgroup then follows from the theorem Intersections of subgroups below, or directly by the same computation with $x$ universally quantified.∎

Exercises for §2.1 →

2.2Intersections, unions, and generated subgroups

Having a supply of subgroups, one asks how they combine. Intersection behaves perfectly; union behaves badly; and the failure of union is what forces the definition of the subgroup generated by a set.

Theorem 2.9 (Intersections of subgroups). Let ${H_{i}}_{i \in I}$ be any nonempty family of subgroups of $G$ . Then

H = i \in I ⋂ H_{i}

is a subgroup of $G$ .

Proof. Each $H_{i}$ contains $e$ , so $e \in H$ and $H \neq = \emptyset$ . Let $a, b \in H$ . Then for every index $i$ we have $a, b \in H_{i}$ , and since $H_{i} \leq G$ the theorem One-step subgroup test gives $a b^{- 1} \in H_{i}$ . As $i$ was arbitrary, $a b^{- 1} \in ⋂_{i} H_{i} = H$ . Applying the theorem One-step subgroup test once more, $H \leq G$ .∎

The proof works for arbitrary, even uncountable, index sets, because the only quantifier manipulation is "for all $i$ ", performed twice. That generality is what makes the next definition legitimate.

Proposition 2.10 (Unions of subgroups are rarely subgroups). Let $H, K \leq G$ . Then $H \cup K \leq G$ if and only if $H \subseteq K$ or $K \subseteq H$ . In particular, taking $G = Z$ , $H = 2 Z$ and $K = 3 Z$ , the union $2 Z \cup 3 Z$ is not a subgroup.

Proof. If one of the two contains the other, the union is the larger of them, hence a subgroup.

Conversely, suppose $H \cup K \leq G$ and that neither containment holds. Choose $h \in H ∖ K$ and $k \in K ∖ H$ . Both lie in $H \cup K$ , so by closure $hk \in H \cup K$ . If $hk \in H$ , then $k = h^{- 1} (hk) \in H$ , contradicting the choice of $k$ . If $hk \in K$ , then $h = (hk) k^{- 1} \in K$ , contradicting the choice of $h$ . Both cases are impossible, so one of the containments must hold.

For the concrete case, $2 \in 2 Z$ and $3 \in 3 Z$ , but $2 + 3 = 5$ lies in neither, so the union is not closed under the operation.∎

Intuition. A subgroup is a closed system: you cannot leave it by multiplying. Gluing two closed systems together at the identity leaves a shape with a hole in the middle — you can step from one branch into the other, and the product lands outside both. Intersection cannot create such a product, because a product of two elements common to every $H_{i}$ is again common to every $H_{i}$ . Union creates one immediately unless one branch was contained in the other all along.

Definition 2.11 (Generated subgroup). Let $S \subseteq G$ . The subgroup generated by $S$ , written $⟨ S ⟩$ , is the intersection of all subgroups of $G$ containing $S$ . If $S = {a}$ we write $⟨ a ⟩$ , and $G$ is cyclic if $G = ⟨ a ⟩$ for some $a$ .

The intersection is over a nonempty family, since $G$ itself contains $S$ , and it is a subgroup by the theorem Intersections of subgroups. By construction $⟨ S ⟩$ is the smallest subgroup containing $S$ : it is contained in every subgroup that contains $S$ . This "intersect everything" definition is clean but non-constructive; the next proposition says what the elements actually are.

Proposition 2.12 (Words describe the generated subgroup). For nonempty $S \subseteq G$ ,

⟨ S ⟩ = {s_{1}^{ε_{1}} s_{2}^{ε_{2}} \dots s_{n}^{ε_{n}} : n \geq 0, s_{i} \in S, ε_{i} \in {1, - 1}},

the set of all finite products of elements of $S$ and their inverses (the empty product being $e$ ). In particular $⟨ a ⟩ = {a^{k} : k \in Z}$ .

Proof. Call the right-hand side $W$ . Every element of $W$ is a product of elements of $S$ and inverses of such, so any subgroup containing $S$ contains all of $W$ ; hence $W \subseteq ⟨ S ⟩$ .

Conversely $W$ is itself a subgroup: it is nonempty (it contains the empty product $e$ ), the concatenation of two words is a word, and the inverse of $s_{1}^{ε_{1}} \dots s_{n}^{ε_{n}}$ is $s_{n}^{- ε_{n}} \dots s_{1}^{- ε_{1}}$ , again a word. It contains $S$ , taking $n = 1$ and $ε_{1} = 1$ . Since $⟨ S ⟩$ is contained in every subgroup containing $S$ , we get $⟨ S ⟩ \subseteq W$ .

The two inclusions give equality. For $S = {a}$ every word collapses by the exponent laws to a single power $a^{k}$ .∎

Definition 2.13 (Order of an element). The order of $a \in G$ , written $ord (a)$ , is the least positive integer $n$ with $a^{n} = e$ , if such an $n$ exists, and $\infty$ otherwise.

Proposition 2.14 (Order equals the size of the generated subgroup). If $ord (a) = n < \infty$ then $⟨ a ⟩ = {e, a, a^{2}, \dots, a^{n - 1}}$ , these $n$ elements are distinct, and $∣ ⟨ a ⟩ ∣ = n$ . Moreover $a^{k} = e$ if and only if $n ∣ k$ .

Proof. By the proposition Words describe the generated subgroup, $⟨ a ⟩ = {a^{k} : k \in Z}$ . Given $k \in Z$ , divide with remainder: $k = q n + r$ with $0 \leq r < n$ . Then $a^{k} = (a^{n})^{q} a^{r} = e^{q} a^{r} = a^{r}$ , so every power equals one of $a^{0}, \dots, a^{n - 1}$ , proving the listed set is all of $⟨ a ⟩$ .

They are distinct: if $a^{i} = a^{j}$ with $0 \leq i < j \leq n - 1$ then $a^{j - i} = e$ with $0 < j - i < n$ , contradicting the minimality in Definition Order of an element.

Finally, if $n ∣ k$ then $a^{k} = (a^{n})^{k / n} = e$ ; and conversely if $a^{k} = e$ , writing $k = q n + r$ as above gives $a^{r} = e$ with $0 \leq r < n$ , which forces $r = 0$ by minimality, i.e. $n ∣ k$ .∎

Example 2.15 (Orders inside a symmetric group). In $S_{4}$ , compute the orders of $σ = (123)$ and $τ = (12) (34)$ , and the size of $⟨ σ ⟩$ .

Solution.

A $3$ -cycle cubed is the identity: $σ^{3} = e$ , and $σ \neq = e$ , $σ^{2} = (132) \neq = e$ . So $ord (σ) = 3$ .
By the proposition Order equals the size of the generated subgroup, $⟨ σ ⟩ = {e, (123), (132)}$ has exactly $3$ elements.
For $τ$ , the two transpositions are disjoint and commute, so $τ^{2} = (12)^{2} (34)^{2} = e$ and $τ \neq = e$ : $ord (τ) = 2$ .
In general the order of a permutation is the least common multiple of its cycle lengths, which gives $3$ and $lcm (2, 2) = 2$ directly.

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Example 2.16 (A subgroup lattice). List all subgroups of $Z /12$ under addition.

Solution.

$Z /12$ is cyclic, generated by $1$ , so every subgroup is cyclic, generated by some $d$ .
The subgroup $⟨ d ⟩$ consists of the multiples of $g cd (d, 12)$ , so the distinct subgroups correspond to the divisors $d \in {1, 2, 3, 4, 6, 12}$ of $12$ .
Explicitly: $⟨ 1 ⟩ = Z /12$ of order $12$ ; $⟨ 2 ⟩$ of order $6$ ; $⟨ 3 ⟩$ of order $4$ ; $⟨ 4 ⟩$ of order $3$ ; $⟨ 6 ⟩$ of order $2$ ; $⟨ 0 ⟩ = {0}$ of order $1$ .
There are exactly $6$ subgroups, one for each divisor of $12$ — a feature of cyclic groups, not of groups in general.

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Figure 2.17. The subgroup lattice of

Z /12

. A line rising from one subgroup to another means containment with nothing in between; the order of each subgroup is the divisor of

12

matching its generator.

Exercises for §2.2 →

2.3Cosets, index, and Lagrange's theorem

A subgroup sits inside a group, and translating it around produces the blocks that measure how much bigger the group is.

Definition 2.18 (Left and right cosets). Let $H \leq G$ and $a \in G$ . The left coset of $H$ by $a$ and the right coset are

a H = {ah : h \in H}, H a = {ha : h \in H} .

The set of all left cosets is written $G / H$ , and the set of right cosets $H \ G$ .

Cosets are subsets of $G$ , not elements of it, and only one of them — namely $eH = H$ — is a subgroup, since any other is a subset missing the identity. The first structural fact is that two cosets are either identical or disjoint, and that they are recognised by a simple membership criterion.

Lemma 2.19 (Coset equality criterion). Let $H \leq G$ and $a, b \in G$ . The following are equivalent.

$a H = b H$ .
$b^{- 1} a \in H$ .
$a \in b H$ .

Consequently, distinct left cosets of $H$ are disjoint.

Proof. $(1) \Rightarrow (3)$ : $a = a e \in a H = b H$ .

$(3) \Rightarrow (2)$ : if $a = bh$ for some $h \in H$ then $b^{- 1} a = h \in H$ .

$(2) \Rightarrow (1)$ : write $b^{- 1} a = h_{0} \in H$ , so $a = b h_{0}$ . Then $a H = b h_{0} H \subseteq b H$ since $h_{0} H \subseteq H$ by closure; and $b = a h_{0}^{- 1}$ gives the reverse inclusion $b H \subseteq a H$ by the same argument. Hence $a H = b H$ .

For disjointness, suppose $a H \cap b H \neq = \emptyset$ and pick $c$ in the intersection. Then $c \in a H$ , so $a H = cH$ by $(3) \Rightarrow (1)$ applied with the roles of $a$ and $c$ exchanged, and likewise $b H = cH$ . Therefore $a H = b H$ : two cosets that meet at all are equal.∎

Remark. The relation $a \sim b ⟺ b^{- 1} a \in H$ is an equivalence relation on $G$ — reflexive because $e \in H$ , symmetric because $H$ is closed under inverses, transitive because $H$ is closed under products — and its equivalence classes are exactly the left cosets. The lemma Coset equality criterion is that statement unpacked, and the partition below is the general fact that equivalence classes partition a set.

Theorem 2.20 (Cosets partition the group into equal-sized blocks). Let $H \leq G$ . Then

every $a \in G$ lies in some left coset, namely $a H$ ;
two left cosets are equal or disjoint;
$∣ a H ∣ = ∣ H ∣$ for every $a \in G$ .

Hence the left cosets of $H$ form a partition of $G$ into blocks each of cardinality $∣ H ∣$ .

Proof. (1) $a = a e \in a H$ since $e \in H$ , so the cosets cover $G$ .

(2) This is the last assertion of the lemma Coset equality criterion.

(3) Define $λ_{a} : H \to a H$ by $λ_{a} (h) = ah$ . It is surjective by the definition of $a H$ . It is injective because $a h_{1} = a h_{2}$ implies $h_{1} = h_{2}$ after multiplying on the left by $a^{- 1}$ — this is cancellation in $G$ , available because $a$ is invertible. So $λ_{a}$ is a bijection and $∣ a H ∣ = ∣ H ∣$ , for infinite $H$ as well as finite.

Covering plus pairwise disjointness is precisely a partition, and (3) says every block has the same size.∎

The equal-size statement deserves emphasis because it is the only place the group axioms are genuinely needed. In an arbitrary magma, translating a subset can shrink it; in a group, left translation by $a$ is a bijection of $G$ with inverse left translation by $a^{- 1}$ , so it cannot. Everything numerical below is that bijection, counted.

Figure 2.21. The four left cosets of

H = ⟨ 4 ⟩ = {0, 4, 8}

Z /12

. The blocks are disjoint, together they exhaust the group, and each contains exactly

∣ H ∣ = 3

elements, so

[Z /12 : H] = 12/3 = 4

Definition 2.22 (Index). The index of $H$ in $G$ , written $[G : H]$ , is the cardinality of the set $G / H$ of left cosets of $H$ in $G$ .

Proposition 2.23 (Left and right cosets are equinumerous). For any $H \leq G$ , the map $a H \mapsto H a^{- 1}$ is a well-defined bijection from $G / H$ to $H \ G$ . Hence the number of left cosets equals the number of right cosets, and $[G : H]$ may be computed with either.

Proof. Well defined. If $a H = b H$ then $b^{- 1} a \in H$ by the lemma Coset equality criterion. Taking inverses, $a^{- 1} b \in H$ , and the right-hand analogue of that lemma gives $H a^{- 1} = H b^{- 1}$ . So the image does not depend on the representative chosen.

Injective. If $H a^{- 1} = H b^{- 1}$ then $a^{- 1} b \in H$ , hence $b^{- 1} a \in H$ , hence $a H = b H$ .

Surjective. Any right coset $Hc$ is the image of $c^{- 1} H$ .∎

The naive map $a H \mapsto H a$ is not well defined in general, and the inverse is doing essential work; this is the first hint that left and right cosets are different objects, a theme that returns with normality.

Theorem 2.24 (Lagrange's theorem). Let $G$ be a finite group and $H \leq G$ . Then

∣ G ∣ = [G : H] ∣ H ∣.

In particular $∣ H ∣$ divides $∣ G ∣$ , and $[G : H] = ∣ G ∣/∣ H ∣$ .

Proof. By the theorem Cosets partition the group into equal-sized blocks, the distinct left cosets of $H$ partition $G$ . Let $n = [G : H]$ be their number, which is finite because the cosets are nonempty and disjoint subsets of the finite set $G$ , and list them as $a_{1} H, \dots, a_{n} H$ . Then

G = a_{1} H \dot{\cup} a_{2} H \dot{\cup} \dots \dot{\cup} a_{n} H,

a disjoint union, so the cardinalities add:

∣ G ∣ = i = 1 \sum n ∣ a_{i} H ∣ = i = 1 \sum n ∣ H ∣ = n ∣ H ∣ = [G : H] ∣ H ∣.

Since $n$ and $∣ H ∣$ are positive integers, $∣ H ∣$ divides $∣ G ∣$ and $n = ∣ G ∣/∣ H ∣$ .∎

Lagrange is a counting argument and nothing more: partition a finite set into blocks of equal size, and the size of the set is the number of blocks times the common block size. The algebra is spent entirely on proving the blocks have equal size, which is why the theorem Cosets partition the group into equal-sized blocks is the real content and Lagrange is its corollary.

Example 2.25 (Index of a subgroup of $S_{4}$ ). Compute $[S_{4} : A_{4}]$ and $[S_{4} : ⟨(1234)⟩]$ .

Solution.

$∣ S_{4} ∣ = 4! = 24$ and $∣ A_{4} ∣ = 12$ , so by Lagrange's theorem $[S_{4} : A_{4}] = 24/12 = 2$ .
The $4$ -cycle $(1234)$ has order $4$ , so by the proposition Order equals the size of the generated subgroup the subgroup it generates has $4$ elements.
Hence $[S_{4} : ⟨(1234)⟩] = 24/4 = 6$ .
Both indices are integers, as Lagrange's theorem forces; an index that failed to be an integer would mean an arithmetic slip somewhere.

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Theorem 2.26 (Tower law for indices). Let $K \leq H \leq G$ . Then

[G : K] = [G : H] [H : K],

with the convention that the identity holds among cardinals, and in particular that $[G : K]$ is finite exactly when both factors are.

Proof. Let ${a_{i}}_{i \in I}$ be a set of representatives for the left cosets of $H$ in $G$ , so $G = ⋃^{˙}_{i} a_{i} H$ with $∣ I ∣ = [G : H]$ , and let ${b_{j}}_{j \in J}$ be representatives for the cosets of $K$ in $H$ , so $H = ⋃^{˙}_{j} b_{j} K$ with $∣ J ∣ = [H : K]$ .

The products cover. Substituting the second decomposition into the first,

G = i \in I ⋃ a_{i} H = i \in I ⋃ j \in J ⋃ a_{i} b_{j} K,

so every element of $G$ lies in some $a_{i} b_{j} K$ .

The products are distinct. Suppose $a_{i} b_{j} K = a_{i^{'}} b_{j^{'}} K$ . Then $a_{i} b_{j}$ and $a_{i^{'}} b_{j^{'}}$ lie in a common coset of $K$ , hence in a common coset of $H$ since $K \subseteq H$ . But $a_{i} b_{j} \in a_{i} H$ and $a_{i^{'}} b_{j^{'}} \in a_{i^{'}} H$ , so $a_{i} H = a_{i^{'}} H$ and therefore $i = i^{'}$ . Cancelling $a_{i}$ , we get $b_{j} K = b_{j^{'}} K$ and so $j = j^{'}$ .

Thus ${a_{i} b_{j} K}_{(i, j) \in I \times J}$ is exactly the set of left cosets of $K$ in $G$ , each appearing once, and $[G : K] = ∣ I \times J ∣ = ∣ I ∣∣ J ∣ = [G : H] [H : K]$ .∎

For finite $G$ the tower law is also immediate by dividing: $\frac{∣ G ∣}{∣ K ∣} = \frac{∣ G ∣}{∣ H ∣} \cdot \frac{∣ H ∣}{∣ K ∣}$ . The proof above is worth having because it needs no finiteness at all — it is a statement about sets of representatives, and it applies to infinite index computations such as $[Z : 12 Z] = [Z : 4 Z] [4 Z : 12 Z] = 4 \cdot 3$ .

Example 2.27 (A tower in $Z$ ). Verify the tower law for $12 Z \leq 4 Z \leq Z$ .

Solution.

$[Z : 4 Z] = 4$ : the cosets are $4 Z, 1 + 4 Z, 2 + 4 Z, 3 + 4 Z$ .
$[4 Z : 12 Z] = 3$ : the cosets are $12 Z, 4 + 12 Z, 8 + 12 Z$ .
$[Z : 12 Z] = 12$ , and indeed $12 = 4 \cdot 3$ , as the theorem Tower law for indices predicts.
Note that Lagrange's theorem itself says nothing here — $Z$ is infinite — while the tower law still applies.

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Exercises for §2.3 →

2.4What Lagrange buys: element orders, Fermat, and Euler

The divisibility statement is worth far more than its one-line proof suggests, because it constrains elements as soon as it is applied to the subgroups they generate.

Corollary 2.28 (The order of an element divides the order of the group). Let $G$ be finite and $a \in G$ . Then $ord (a)$ divides $∣ G ∣$ , and $a^{∣ G ∣} = e$ .

Proof. Let $n = ord (a)$ , which is finite since the powers of $a$ cannot all be distinct in a finite group. By the proposition Order equals the size of the generated subgroup, $∣ ⟨ a ⟩ ∣ = n$ , and $⟨ a ⟩ \leq G$ , so Lagrange's theorem gives $n ∣ ∣ G ∣$ .

Write $∣ G ∣ = nk$ . Then $a^{∣ G ∣} = (a^{n})^{k} = e^{k} = e$ .∎

Corollary 2.29 (Groups of prime order are cyclic). If $∣ G ∣ = p$ with $p$ prime, then $G ≅ Z / p$ , and $G$ has no subgroups other than ${e}$ and $G$ .

Proof. Since $p \geq 2$ , choose $a \in G$ with $a \neq = e$ . By Lagrange's theorem $∣ ⟨ a ⟩ ∣$ divides $p$ , so it is $1$ or $p$ . It is not $1$ , because $⟨ a ⟩$ contains the two distinct elements $e$ and $a$ . Hence $∣ ⟨ a ⟩ ∣ = p = ∣ G ∣$ and $G = ⟨ a ⟩$ is cyclic of order $p$ , so $G ≅ Z / p$ by the classification of cyclic groups in Groups.

The same divisibility argument applied to an arbitrary $H \leq G$ gives $∣ H ∣ \in {1, p}$ , i.e. $H = {e}$ or $H = G$ .∎

This is the first classification theorem in the subject: there is exactly one group of each prime order, up to isomorphism. It also explains why $p$ , $p^{2}$ , $pq$ and so on are the interesting orders — the constraints weaken as the number of prime factors grows.

Corollary 2.30 (Euler's theorem). Let $n \geq 1$ and let $a$ be an integer with $g cd (a, n) = 1$ . Then

a^{φ (n)} \equiv 1 (mod n),

where $φ$ is Euler's totient function.

Proof. Let $G = (Z / n)^{\times}$ , the set of residue classes mod $n$ that are coprime to $n$ , under multiplication. This is a group: it is closed because a product of two units is a unit, the class of $1$ is the identity, and each unit has an inverse by Bézout's identity, as established in Modular Arithmetic. By definition of the totient function, $∣ G ∣ = φ (n)$ .

Since $g cd (a, n) = 1$ , the class $\overset{a}{ˉ}$ lies in $G$ . The corollary The order of an element divides the order of the group gives $\overset{a}{ˉ}^{∣ G ∣} = \overset{ˉ}{1}$ , that is, $\overset{a}{ˉ}^{φ (n)} = \overset{ˉ}{1}$ , which is the congruence claimed.∎

Corollary 2.31 (Fermat's little theorem). Let $p$ be prime and $a$ an integer not divisible by $p$ . Then $a^{p - 1} \equiv 1 (mod p)$ , and for every integer $a$ , $a^{p} \equiv a (mod p)$ .

Proof. For prime $p$ , every one of $1, 2, \dots, p - 1$ is coprime to $p$ , so $φ (p) = p - 1$ . If $p ∤ a$ then $g cd (a, p) = 1$ and the corollary Euler's theorem gives $a^{p - 1} \equiv 1$ .

Multiplying by $a$ gives $a^{p} \equiv a (mod p)$ when $p ∤ a$ ; and if $p ∣ a$ then both sides are $\equiv 0$ , so $a^{p} \equiv a$ holds for every integer $a$ .∎

Intuition. Fermat's little theorem is usually presented as a fact about remainders, and it looks like a coincidence of arithmetic. Lagrange explains it: the nonzero residues mod $p$ form a group of size $p - 1$ , and every element of every finite group is killed by the group's order. The number $p - 1$ in the exponent is not arithmetic at all — it is the size of a symmetry group, and the theorem is the statement that a symmetry repeated as many times as there are symmetries returns you to where you started.

Example 2.32 (Computing a large power modulo 100). Find the last two digits of $7^{2024}$ .

Solution.

Last two digits means computing modulo $100$ . Since $g cd (7, 100) = 1$ , the corollary Euler's theorem applies.
$φ (100) = 100 (1 - \frac{1}{2}) (1 - \frac{1}{5}) = 40$ , so $7^{40} \equiv 1 (mod 100)$ .
$2024 = 40 \cdot 50 + 24$ , so $7^{2024} \equiv 7^{24} (mod 100)$ .
$7^{2} = 49$ , $7^{4} = 4 9^{2} = 2401 \equiv 1 (mod 100)$ . So $7^{24} = (7^{4})^{6} \equiv 1$ .
The last two digits are $01$ . Note the actual order of $7$ is $4$ , a proper divisor of $φ (100) = 40$ — Euler's exponent always works but is not always least.

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Example 2.33 (Constraining subgroup orders). A group $G$ has order $30$ . Which of $4, 5, 6, 7, 15$ can be the order of a subgroup, on the evidence of Lagrange alone?

Solution.

By Lagrange's theorem, the order of any subgroup divides $30 = 2 \cdot 3 \cdot 5$ .
The divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$ .
So $4$ and $7$ are impossible, while $5$ , $6$ and $15$ are not excluded.
"Not excluded" is not "exists": Lagrange is a one-way implication. For order $30$ all three do occur, but the converse of Lagrange is false in general — see the pitfall below.

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Pitfall. The converse of Lagrange is false. $A_{4}$ has order $12$ , and $6$ divides $12$ , yet $A_{4}$ has no subgroup of order $6$ . A subgroup of index $2$ would be normal by the proposition Subgroups of index two are normal, and would have to contain every square; but the eight $3$ -cycles of $A_{4}$ are all squares (each equals the square of its own inverse), and $8 > 6$ . So " $d$ divides $∣ G ∣$ " never by itself produces a subgroup of order $d$ . The Sylow theorems below recover the statement for prime powers, and only for prime powers.

Proposition 2.34 (Groups of order $p^{2}$ and small orders). Let $∣ G ∣ = p^{2}$ with $p$ prime. Then every element of $G$ has order $1$ , $p$ or $p^{2}$ , and if some element has order $p^{2}$ then $G$ is cyclic.

Proof. By the corollary The order of an element divides the order of the group, the order of $a \in G$ divides $p^{2}$ , and the divisors of $p^{2}$ are $1$ , $p$ and $p^{2}$ . Order $1$ happens only for $a = e$ .

If $ord (a) = p^{2}$ then $∣ ⟨ a ⟩ ∣ = p^{2} = ∣ G ∣$ by the proposition Order equals the size of the generated subgroup, so $⟨ a ⟩ = G$ and $G$ is cyclic.∎

Exercises for §2.4 →

2.5Normal subgroups and quotient groups

We now try to make $G / H$ — the set of cosets — into a group. The only plausible rule is to multiply representatives, and that rule works for some subgroups and fails for others. Understanding exactly when is the central point of the chapter.

Definition 2.35 (Normal subgroup). A subgroup $N \leq G$ is normal, written $N ⊴ G$ , if

g N g^{- 1} = N for every g \in G,

where $g N g^{- 1} = {g n g^{- 1} : n \in N}$ .

Theorem 2.36 (Characterisations of normality). For $N \leq G$ the following are equivalent.

$g N g^{- 1} = N$ for all $g \in G$ .
$g N g^{- 1} \subseteq N$ for all $g \in G$ .
$g N = N g$ for all $g \in G$ — the left and right coset partitions coincide.
Every left coset of $N$ is a right coset of $N$ .

Proof. $(1) \Rightarrow (2)$ is trivial.

$(2) \Rightarrow (1)$ . Fix $g$ . Applying (2) to $g^{- 1}$ gives $g^{- 1} N g \subseteq N$ ; conjugating this containment by $g$ gives $N \subseteq g N g^{- 1}$ . Together with (2) this yields equality. This step is the reason (2) suffices: the hypothesis is quantified over all $g$ , so it may be applied to $g^{- 1}$ as well, and one-sided containment for all $g$ upgrades to equality.

$(1) \Rightarrow (3)$ . Multiply $g N g^{- 1} = N$ on the right by $g$ : $g N = N g$ .

$(3) \Rightarrow (4)$ is trivial.

$(4) \Rightarrow (3)$ . Suppose $g N = N h$ for some $h$ . The element $g = g e$ lies in $g N$ , hence in $N h$ , so $g h^{- 1} \in N$ and therefore $N h = N g$ by the right-coset version of the lemma Coset equality criterion. Thus $g N = N g$ .

$(3) \Rightarrow (1)$ . Multiply $g N = N g$ on the right by $g^{- 1}$ .∎

Condition (4) is worth reading twice. It does not say that the left coset $g N$ equals the right coset $N g$ — it says only that it equals some right coset — and the proof shows that "some" is automatically "that one", because a coset is determined by any element it contains.

Pitfall. Normality does not mean $g n = n g$ for each $n \in N$ . It means $g n g^{- 1}$ lands back in $N$ , possibly as a different element of $N$ . The subgroup of $S_{3}$ generated by a $3$ -cycle is normal, yet no transposition commutes with a $3$ -cycle: conjugation permutes ${(123), (132)}$ by swapping them. Confusing the two conditions is the difference between normal and central.

Example 2.37 (Normal and non-normal subgroups of $S_{3}$ ). In $S_{3}$ , decide whether $A_{3} = ⟨(123)⟩$ and $⟨(12)⟩$ are normal.

Solution.

$∣ A_{3} ∣ = 3$ and $∣ S_{3} ∣ = 6$ , so $[S_{3} : A_{3}] = 2$ and $A_{3}$ is normal by the proposition Subgroups of index two are normal below.
Directly: conjugation preserves cycle type, so a conjugate of a $3$ -cycle is a $3$ -cycle, and the only $3$ -cycles are the two in $A_{3}$ . Hence $σ A_{3} σ^{- 1} = A_{3}$ .
For $H = {e, (12)}$ , conjugate by $g = (13)$ : $(13) (12) (13)^{- 1} = (32) = (23) \in / H$ .
So $H$ is not normal. Its left coset $(13) H = {(13), (13) (12)} = {(13), (123)}$ differs from $H (13) = {(13), (132)}$ , exactly as condition (3) of the theorem Characterisations of normality predicts.

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Proposition 2.38 (Subgroups of index two are normal). If $H \leq G$ and $[G : H] = 2$ , then $H ⊴ G$ .

Proof. There are exactly two left cosets. One is $H$ itself, and since the cosets partition $G$ by the theorem Cosets partition the group into equal-sized blocks, the other must be $G ∖ H$ . The same argument applies to right cosets: they are $H$ and $G ∖ H$ .

Now take $g \in G$ . If $g \in H$ then $g H = H = H g$ . If $g \in / H$ then $g H$ is the coset different from $H$ , so $g H = G ∖ H$ , and likewise $H g = G ∖ H$ . In both cases $g H = H g$ , so $H ⊴ G$ by the theorem Characterisations of normality.∎

Proposition 2.39 (Every subgroup of an abelian group is normal; the centre is normal). If $G$ is abelian then every $H \leq G$ is normal. For any group $G$ , $Z (G) ⊴ G$ .

Proof. If $G$ is abelian then $g H g^{- 1} = {g h g^{- 1}} = {h g g^{- 1}} = H$ for every $g$ .

For the centre, let $z \in Z (G)$ and $g \in G$ . Then $g z g^{- 1} = z g g^{- 1} = z \in Z (G)$ , so $g Z (G) g^{- 1} \subseteq Z (G)$ for every $g$ , and condition (2) of the theorem Characterisations of normality gives normality.∎

We can now attempt the construction. Define a product on the set $G / N$ of left cosets by

(a N) (b N) = ab N .

The formula refers to $a$ and $b$ , but the inputs are the cosets $a N$ and $b N$ , and a coset has many representatives — $∣ N ∣$ of them. So there is something to check before this is a definition at all: if $a N = a^{'} N$ and $b N = b^{'} N$ , does $ab N = a^{'} b^{'} N$ ? Nothing in the formula guarantees it, and for a non-normal subgroup it is false.

Theorem 2.40 (The quotient group exists exactly when $N$ is normal). Let $N \leq G$ . The rule $(a N) (b N) = ab N$ is well defined on $G / N$ if and only if $N ⊴ G$ . In that case $G / N$ is a group under this operation, with identity $N = e N$ and $(a N)^{- 1} = a^{- 1} N$ ; it is called the quotient group of $G$ by $N$ , and for finite $G$ , $∣ G / N ∣ = [G : N] = ∣ G ∣/∣ N ∣$ .

Proof. Normal $\Rightarrow$ well defined. Suppose $N ⊴ G$ , $a N = a^{'} N$ and $b N = b^{'} N$ . By the lemma Coset equality criterion, $a^{'} = a n_{1}$ and $b^{'} = b n_{2}$ with $n_{1}, n_{2} \in N$ . Then

a^{'} b^{'} = a n_{1} b n_{2} = ab (b^{- 1} n_{1} b) n_{2} .

By normality, $b^{- 1} n_{1} b \in b^{- 1} N b = N$ , so $(b^{- 1} n_{1} b) n_{2} \in N$ by closure. Hence $a^{'} b^{'} \in ab N$ , and the lemma Coset equality criterion gives $a^{'} b^{'} N = ab N$ . The product does not depend on the representatives, so it is a genuine operation on $G / N$ .

Well defined $\Rightarrow$ normal. Suppose the rule is well defined. Fix $g \in G$ and $n \in N$ . Since $n \in N$ , the lemma Coset equality criterion gives $n N = N = e N$ . Feeding the two equal pairs of inputs $(n N, g N)$ and $(e N, g N)$ into the rule, the outputs must agree:

n g N = (n N) (g N) = (e N) (g N) = e g N = g N .

By the lemma Coset equality criterion applied to $n g N = g N$ , we get $g^{- 1} (n g) \in N$ , that is $g^{- 1} n g \in N$ . As $g \in G$ and $n \in N$ were arbitrary, $g^{- 1} N g \subseteq N$ for every $g$ , and condition (2) of the theorem Characterisations of normality gives $N ⊴ G$ .

Group axioms. Associativity: $((a N) (b N)) (c N) = (ab) c N = a (b c) N = (a N) ((b N) (c N))$ , inherited from $G$ . Identity: $(e N) (a N) = a N = (a N) (e N)$ . Inverses: $(a N) (a^{- 1} N) = a a^{- 1} N = e N$ . So $G / N$ is a group, and its underlying set is the set of cosets, of size $[G : N]$ , which equals $∣ G ∣/∣ N ∣$ by Lagrange's theorem when $G$ is finite.∎

Remark. The well-definedness check is the whole theorem. Every later construction that "mods out" — quotient rings by ideals in Ideals and Quotient Rings, quotient vector spaces, quotient topologies — repeats this same two-line argument in its own setting, and in each case the special class of sub-objects (normal subgroup, two-sided ideal) is defined precisely so that the check passes. If you remember one computation from this chapter, make it $a^{'} b^{'} = ab (b^{- 1} n_{1} b) n_{2}$ .

Example 2.41 (Failure of well-definedness for a non-normal subgroup). Let $G = S_{3}$ and $H = {e, (12)}$ . Show that $(a H) (b H) = ab H$ is not well defined.

Solution.

Take $a = (13)$ and $a^{'} = (13) (12) = (123)$ . Since $a^{'} = ah$ with $h = (12) \in H$ , we have $a H = a^{'} H$ .
Take $b = (13)$ and $b^{'} = b$ , so certainly $b H = b^{'} H$ .
Then $ab = (13) (13) = e$ , so $ab H = H = {e, (12)}$ .
But $a^{'} b^{'} = (123) (13)$ . Composing right to left: $1 \mapsto 3 \mapsto 1$ , $2 \mapsto 2 \mapsto 3$ , $3 \mapsto 1 \mapsto 2$ , so $a^{'} b^{'} = (23)$ and $a^{'} b^{'} H = {(23), (23) (12)} = {(23), (132)}$ .
The two answers ${e, (12)}$ and ${(23), (132)}$ are different cosets, so the "product" depends on the representatives and is not an operation. Consistently with the theorem The quotient group exists exactly when $N$ is normal, $H$ is not normal in $S_{3}$ .

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Example 2.42 (The quotient $Z / n Z$ ). Identify the quotient of $(Z, +)$ by $n Z$ .

Solution.

$Z$ is abelian, so $n Z ⊴ Z$ by the proposition Every subgroup of an abelian group is normal; the centre is normal, and the quotient exists.
The cosets are $k + n Z$ for $k = 0, 1, \dots, n - 1$ , and these are distinct by the lemma Coset equality criterion: $k + n Z = l + n Z$ iff $n ∣ k - l$ .
So $[Z : n Z] = n$ and $∣ Z / n Z ∣ = n$ .
The operation $(k + n Z) + (l + n Z) = (k + l) + n Z$ is addition modulo $n$ . Modular arithmetic is therefore not a separate invention: it is the quotient construction applied to the simplest infinite group.

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Proposition 2.43 (Kernels are normal, and every normal subgroup is a kernel). Let $φ : G \to K$ be a group homomorphism. Then $ker φ = {g \in G : φ (g) = e_{K}}$ is a normal subgroup of $G$ . Conversely, if $N ⊴ G$ then the map $π : G \to G / N$ , $π (g) = g N$ , is a surjective homomorphism with $ker π = N$ .

Proof. Kernels are normal. $ker φ$ is a subgroup: it contains $e$ since $φ (e) = e_{K}$ , and if $φ (a) = φ (b) = e_{K}$ then $φ (a b^{- 1}) = φ (a) φ (b)^{- 1} = e_{K}$ , so the theorem One-step subgroup test applies. For normality, let $n \in ker φ$ and $g \in G$ ; then

φ (g n g^{- 1}) = φ (g) φ (n) φ (g)^{- 1} = φ (g) e_{K} φ (g)^{- 1} = e_{K},

so $g n g^{- 1} \in ker φ$ , and condition (2) of the theorem Characterisations of normality gives $ker φ ⊴ G$ .

Every normal subgroup is a kernel. Let $N ⊴ G$ and define $π (g) = g N$ ; this is legitimate by the theorem The quotient group exists exactly when $N$ is normal. It is a homomorphism because $π (ab) = ab N = (a N) (b N) = π (a) π (b)$ , precisely by the definition of the quotient operation, and it is surjective since every coset is of the form $g N$ . Finally $π (g) = e N$ iff $g N = N$ iff $g \in N$ by the lemma Coset equality criterion, so $ker π = N$ .∎

So "normal subgroup" and "kernel of a homomorphism" are the same class of objects described two ways: internally, by stability under conjugation, and externally, by being the collapse of a map. This equivalence is what the first isomorphism theorem in Group Homomorphisms and Isomorphisms exploits, and it is the reason normality is the right hypothesis rather than an arbitrary one.

Definition 2.44 (Simple group). A group $G$ is simple if $∣ G ∣ > 1$ and its only normal subgroups are ${e}$ and $G$ .

By the corollary Groups of prime order are cyclic, every group of prime order is simple, since it has no proper nontrivial subgroups at all. Simple groups are the objects that cannot be broken down by the quotient construction — if $N$ is a proper nontrivial normal subgroup, $G$ is assembled from the strictly smaller groups $N$ and $G / N$ — which is why classifying them was the central project of twentieth-century group theory.

Example 2.45 (Order of a quotient). Let $G = Z /24$ and $N = ⟨ 8 ⟩$ . Compute $∣ N ∣$ and $∣ G / N ∣$ , and identify $G / N$ .

Solution.

$⟨ 8 ⟩ = {0, 8, 16}$ , so $∣ N ∣ = 3$ ; equivalently $ord (8) = 24/ g cd (8, 24) = 24/8 = 3$ .
$G$ is abelian so $N$ is normal, and by the theorem The quotient group exists exactly when $N$ is normal, $∣ G / N ∣ = 24/3 = 8$ .
The coset $1 + N$ generates: its multiples run through all $8$ cosets before repeating, since $k \in N$ first happens at $k = 8$ .
Hence $G / N ≅ Z /8$ . In general $(Z / mn) / ⟨ m ⟩ ≅ Z / m$ .

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Exercises for §2.5 →

2.6Conjugacy, actions, and the class equation

The Sylow theorems are proved by letting a group act on a set built out of the group itself and counting orbits. This section assembles the two counting facts that argument needs.

Definition 2.46 (Group action, orbit, stabiliser). An action of $G$ on a set $X$ is a map $G \times X \to X$ , $(g, x) \mapsto g \cdot x$ , with $e \cdot x = x$ and $g \cdot (h \cdot x) = (g h) \cdot x$ . The orbit of $x$ is $O_{x} = {g \cdot x : g \in G}$ and the stabiliser is $G_{x} = {g \in G : g \cdot x = x}$ .

Theorem 2.47 (Orbit–stabiliser theorem). Let $G$ act on $X$ and $x \in X$ . Then $G_{x} \leq G$ and the map $g G_{x} \mapsto g \cdot x$ is a bijection from $G / G_{x}$ onto $O_{x}$ . Hence

∣ O_{x} ∣ = [G : G_{x}], and for finite G, ∣ O_{x} ∣ = \frac{∣ G ∣}{∣ G _{x} ∣} .

Proof. $G_{x}$ is a subgroup: $e \in G_{x}$ , and if $g, h \in G_{x}$ then $(g h^{- 1}) \cdot x = g \cdot (h^{- 1} \cdot x) = g \cdot x = x$ , using that $h \cdot x = x$ implies $h^{- 1} \cdot x = x$ .

Well defined and injective. $g G_{x} = h G_{x} ⟺ h^{- 1} g \in G_{x} ⟺ (h^{- 1} g) \cdot x = x ⟺ g \cdot x = h \cdot x$ . Reading the chain left to right gives well-definedness, right to left injectivity.

Surjective. Every element of $O_{x}$ is $g \cdot x$ for some $g$ , the image of $g G_{x}$ .

So $∣ O_{x} ∣ = ∣ G / G_{x} ∣ = [G : G_{x}]$ , and Lagrange's theorem converts this to $∣ G ∣/∣ G_{x} ∣$ in the finite case.∎

In particular every orbit size divides $∣ G ∣$ — a fact that does most of the work in Sylow arguments, because it limits the possible orbit sizes to a short list before any group-specific information is used.

Theorem 2.48 (The class equation). Let $G$ be a finite group acting on itself by conjugation, $g \cdot x = gx g^{- 1}$ . The orbits are the conjugacy classes, the stabiliser of $x$ is $C_{G} (x)$ , and

∣ G ∣ = ∣ Z (G) ∣ + i = 1 \sum r [G : C_{G} (x_{i})],

where $x_{1}, \dots, x_{r}$ are representatives of the conjugacy classes with more than one element.

Proof. Conjugation is an action: $e x e^{- 1} = x$ and $g (h x h^{- 1}) g^{- 1} = (g h) x (g h)^{- 1}$ . Its stabilisers are ${g : gx g^{- 1} = x} = C_{G} (x)$ by Definition Centre and centraliser.

The orbits partition $G$ , so $∣ G ∣$ is the sum of the orbit sizes. An orbit is a single point exactly when $gx g^{- 1} = x$ for every $g$ , i.e. $x \in Z (G)$ ; those orbits contribute $∣ Z (G) ∣$ in total. Each remaining orbit has size $[G : C_{G} (x_{i})] > 1$ by the theorem Orbit–stabiliser theorem, and summing gives the displayed identity.∎

Corollary 2.49 (Nontrivial $p$ -groups have nontrivial centre). If $∣ G ∣ = p^{n}$ with $p$ prime and $n \geq 1$ , then $Z (G) \neq = {e}$ .

Proof. Apply the theorem The class equation. Each index $[G : C_{G} (x_{i})]$ divides $∣ G ∣ = p^{n}$ and exceeds $1$ , so each is a positive power of $p$ and in particular is divisible by $p$ . Also $p$ divides $∣ G ∣$ . Rearranging the class equation,

∣ Z (G) ∣ = ∣ G ∣ - i = 1 \sum r [G : C_{G} (x_{i})],

and $p$ divides both terms on the right, so $p ∣ ∣ Z (G) ∣$ . Since $∣ Z (G) ∣ \geq 1$ (it contains $e$ ), we conclude $∣ Z (G) ∣ \geq p > 1$ .∎

Example 2.50 (Groups of order $p^{2}$ are abelian). Use the centre to show that every group of order $p^{2}$ is abelian.

Solution.

By the corollary Nontrivial $p$ -groups have nontrivial centre, $∣ Z (G) ∣ \in {p, p^{2}}$ .
Suppose $∣ Z (G) ∣ = p$ . Then $G / Z (G)$ has order $p$ , hence is cyclic by the corollary Groups of prime order are cyclic; say it is generated by $g Z (G)$ .
Then every element of $G$ has the form $g^{k} z$ with $z \in Z (G)$ , and two such elements commute: $(g^{j} z_{1}) (g^{k} z_{2}) = g^{j + k} z_{1} z_{2} = (g^{k} z_{2}) (g^{j} z_{1})$ .
So $G$ is abelian, forcing $Z (G) = G$ and $∣ Z (G) ∣ = p^{2}$ , contradicting the assumption. Hence $∣ Z (G) ∣ = p^{2}$ and $G$ is abelian all along.

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Exercises for §2.6 →

2.7The Sylow theorems

Lagrange forbids subgroups of orders that do not divide $∣ G ∣$ ; the pitfall after the example Constraining subgroup orders shows it cannot guarantee them. Sylow's theorems give the sharpest true converse: prime powers are always realised, and the realisations are unique up to conjugacy and counted by congruences.

Definition 2.51 (Sylow $p$ -subgroup). Let $∣ G ∣ = p^{a} m$ with $p$ prime, $a \geq 1$ and $p ∤ m$ . A subgroup of $G$ of order exactly $p^{a}$ is a Sylow $p$ -subgroup. The set of them is $Syl_{p} (G)$ and its size is written $n_{p}$ .

Theorem 2.52 (Sylow I: existence). Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ ; that is, $n_{p} \geq 1$ .

Proof. We use Wielandt's counting argument. Let $X$ be the collection of all subsets of $G$ of size $p^{a}$ , and let $G$ act on $X$ by left translation, $g \cdot S = g S$ . This is an action: $e S = S$ and $g (h S) = (g h) S$ , and $∣ g S ∣ = ∣ S ∣$ because left translation is a bijection, so $X$ is preserved.

The size of $X$ is not divisible by $p$ . We have $∣ X ∣ = (p ^{a} p ^{a} m)$ . Writing the binomial coefficient as $\prod_{k = 0}^{p^{a} - 1} \frac{p ^{a} m - k}{p ^{a} - k}$ , compare the power of $p$ dividing numerator and denominator of each factor: for $0 < k < p^{a}$ , if $p^{j}$ exactly divides $k$ then $j < a$ , so $p^{j}$ exactly divides both $p^{a} m - k$ and $p^{a} - k$ . The factor $k = 0$ contributes $p^{a} m / p^{a} = m$ , which is prime to $p$ . Hence every power of $p$ cancels and $p ∤ ∣ X ∣$ .

Extracting a subgroup. The orbits partition $X$ , so if $p$ divided every orbit size it would divide $∣ X ∣$ . It does not, so some orbit $O_{S}$ has $p ∤ ∣ O_{S} ∣$ . Let $H = G_{S}$ be the stabiliser of that $S$ . By the theorem Orbit–stabiliser theorem, $∣ O_{S} ∣ = ∣ G ∣/∣ H ∣ = p^{a} m /∣ H ∣$ , and since $p^{a}$ cannot divide $∣ O_{S} ∣$ , we get $p^{a} ∣ ∣ H ∣$ , so $∣ H ∣ \geq p^{a}$ .

The other inequality. Fix $s \in S$ . For $h \in H$ we have $h S = S$ , so $h s \in S$ ; the map $h \mapsto h s$ is injective, so $∣ H ∣ \leq ∣ S ∣ = p^{a}$ .

Therefore $∣ H ∣ = p^{a}$ and $H$ is a Sylow $p$ -subgroup.∎

The two Sylow theorems that remain are proved by the same technique — let $G$ , or a $p$ -subgroup of it, act on the set of Sylow $p$ -subgroups by conjugation, and read off the orbit sizes. We state them in full and give the group-action argument in outline, assuming only the orbit counting already proved and one standard lemma: if a $p$ -group $P$ acts on a finite set $Y$ , then $∣ Y ∣ \equiv ∣ Y^{P} ∣ (mod p)$ , where $Y^{P}$ is the set of fixed points. That lemma follows immediately from the theorem Orbit–stabiliser theorem, since every non-fixed orbit has size a positive power of $p$ .

Theorem 2.53 (Sylow II: conjugacy and containment). Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then all Sylow $p$ -subgroups of $G$ are conjugate: if $P, Q \in Syl_{p} (G)$ there is $g \in G$ with $g P g^{- 1} = Q$ . Moreover every $p$ -subgroup of $G$ is contained in some Sylow $p$ -subgroup.

Proof. Sketch, by the action of a $p$ -subgroup on cosets. Let $P \in Syl_{p} (G)$ and let $Q$ be any $p$ -subgroup. Let $Q$ act on the coset space $G / P$ by $q \cdot (g P) = q g P$ , which is an action since left multiplication respects cosets.

By the fixed-point lemma above, $∣ G / P ∣ \equiv ∣ (G / P)^{Q} ∣ (mod p)$ . Now $∣ G / P ∣ = m$ is prime to $p$ , so $(G / P)^{Q}$ is nonempty: there is a coset $g P$ with $q g P = g P$ for all $q \in Q$ . That says $g^{- 1} Q g \subseteq P$ , i.e. $Q \subseteq g P g^{- 1}$ , and $g P g^{- 1}$ is a Sylow $p$ -subgroup because conjugation is an automorphism and preserves order. This proves the containment statement.

If moreover $Q$ is itself Sylow, then $∣ Q ∣ = ∣ g P g^{- 1} ∣ = p^{a}$ and the containment $Q \subseteq g P g^{- 1}$ forces equality, giving conjugacy.∎

Theorem 2.54 (Sylow III: counting). Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ , and let $n_{p} = ∣ Syl_{p} (G) ∣$ . Then

n_{p} \equiv 1 (mod p), n_{p} ∣ m, n_{p} = [G : N_{G} (P)]

for any $P \in Syl_{p} (G)$ , where $N_{G} (P) = {g : g P g^{- 1} = P}$ is the normaliser of $P$ .

Proof. Sketch, by two conjugation actions. Let $G$ act on $Syl_{p} (G)$ by conjugation. By the theorem Sylow II: conjugacy and containment this action is transitive, so there is one orbit and the theorem Orbit–stabiliser theorem gives $n_{p} = [G : N_{G} (P)]$ , the stabiliser of $P$ being exactly its normaliser. Since $P \leq N_{G} (P)$ , the theorem Tower law for indices gives $[G : N_{G} (P)] ∣ [G : P] = m$ , which is the divisibility statement.

Now restrict the action to $P$ acting on $Syl_{p} (G)$ by conjugation. By the fixed-point lemma, $n_{p} \equiv ∣ Syl_{p} (G)^{P} ∣ (mod p)$ . A fixed point is a $Q$ with $PQ P^{- 1}$ -stability, i.e. $P \leq N_{G} (Q)$ ; then $P$ and $Q$ are both Sylow $p$ -subgroups of $N_{G} (Q)$ , in which $Q$ is normal, so the theorem Sylow II: conjugacy and containment applied inside $N_{G} (Q)$ forces $P = Q$ . Hence the only fixed point is $P$ itself and $n_{p} \equiv 1 (mod p)$ .∎

Remark. What the sketches assume is exactly two things: the theorem Orbit–stabiliser theorem, proved above, and the fixed-point congruence $∣ Y ∣ \equiv ∣ Y^{P} ∣ (mod p)$ for a $p$ -group acting on a finite set, which is that theorem applied orbit by orbit. No further machinery is used, and the argument for Sylow I above is complete rather than a sketch.

Corollary 2.55 (A unique Sylow subgroup is normal). Let $P \in Syl_{p} (G)$ . Then $P ⊴ G$ if and only if $n_{p} = 1$ .

Proof. If $n_{p} = 1$ then $P$ is the only Sylow $p$ -subgroup. For any $g \in G$ , $g P g^{- 1}$ has order $p^{a}$ and is therefore also a Sylow $p$ -subgroup, so $g P g^{- 1} = P$ ; thus $P ⊴ G$ .

Conversely, if $P ⊴ G$ and $Q \in Syl_{p} (G)$ , the theorem Sylow II: conjugacy and containment gives $Q = g P g^{- 1} = P$ for some $g$ , so $P$ is the unique Sylow $p$ -subgroup and $n_{p} = 1$ .∎

Figure 2.56. Sylow III for

p = 3

in a group of order

12 = 3 \cdot 4

: the count

n_{3}

must divide

m = 4

and be congruent to

1

modulo

3

. Of the three divisors of

4

, only

1

and

4

survive both conditions.

Exercises for §2.7 →

2.8Sylow in practice: counting and non-simplicity

The three theorems become a tool the moment one notices that the two conditions on $n_{p}$ are usually satisfied by only one number.

Method 2.57 (Locating a normal Sylow subgroup). To show a finite group $G$ of order $n$ is not simple:

Factor $n = p^{a} m$ for a chosen prime $p$ with $p ∤ m$ .
List the divisors of $m$ .
Keep those congruent to $1$ modulo $p$ ; these are the possible values of $n_{p}$ by the theorem Sylow III: counting.
If only $n_{p} = 1$ survives, the Sylow $p$ -subgroup is normal by the corollary A unique Sylow subgroup is normal, and $G$ is not simple.
If several values survive, count elements: distinct Sylow $p$ -subgroups for different primes intersect trivially, and the element counts often overflow $∣ G ∣$ unless some $n_{p} = 1$ .

Theorem 2.58 (No group of order $pq$ is simple). Let $p < q$ be primes and $∣ G ∣ = pq$ . Then $G$ has a normal Sylow $q$ -subgroup; in particular $G$ is not simple.

Proof. Here $∣ G ∣ = q^{1} p$ with $q ∤ p$ , so by the theorem Sylow III: counting, $n_{q} ∣ p$ and $n_{q} \equiv 1 (mod q)$ . The divisors of the prime $p$ are $1$ and $p$ , so $n_{q} \in {1, p}$ .

Suppose $n_{q} = p$ . Then $p \equiv 1 (mod q)$ , so $q ∣ p - 1$ and therefore $q \leq p - 1 < p$ . This contradicts $p < q$ .

Hence $n_{q} = 1$ , and the unique Sylow $q$ -subgroup is normal by the corollary A unique Sylow subgroup is normal. It has order $q$ , so it is neither trivial nor all of $G$ , and $G$ is not simple.∎

Example 2.59 (Groups of order 15 are cyclic). Show that every group of order $15$ is isomorphic to $Z /15$ .

Solution.

$15 = 3 \cdot 5$ . By the theorem Sylow III: counting, $n_{5} ∣ 3$ and $n_{5} \equiv 1 (mod 5)$ , so $n_{5} = 1$ ; and $n_{3} ∣ 5$ with $n_{3} \equiv 1 (mod 3)$ , so $n_{3} = 1$ .
Let $P$ of order $3$ and $Q$ of order $5$ be the unique Sylow subgroups; both are normal by the corollary A unique Sylow subgroup is normal.
$P \cap Q$ is a subgroup whose order divides both $3$ and $5$ by Lagrange's theorem, so $P \cap Q = {e}$ . Hence elements of $P$ and $Q$ commute and $PQ$ has $15$ elements, so $G ≅ P \times Q ≅ Z /3 \times Z /5$ .
Since $g cd (3, 5) = 1$ , that product is cyclic of order $15$ . So $G ≅ Z /15$ , and there is exactly one group of order $15$ .

□

Example 2.60 (Counting Sylow subgroups of $S_{4}$ ). Determine $n_{3}$ and $n_{2}$ for $S_{4}$ .

Solution.

$∣ S_{4} ∣ = 24 = 2^{3} \cdot 3$ . For $p = 3$ : $n_{3} ∣ 8$ and $n_{3} \equiv 1 (mod 3)$ , so $n_{3} \in {1, 4}$ .
$S_{4}$ contains eight $3$ -cycles, and each Sylow $3$ -subgroup has order $3$ , hence contains exactly two of them. Distinct subgroups of prime order meet trivially, so $n_{3} = 8/2 = 4$ .
For $p = 2$ : $n_{2} ∣ 3$ and $n_{2} \equiv 1 (mod 2)$ , so $n_{2} \in {1, 3}$ . A Sylow $2$ -subgroup has order $8$ ; one of them is the dihedral group of symmetries of the square $⟨(1234), (13)⟩$ , and it is not normal since conjugating by $(12)$ gives a different subgroup.
Hence $n_{2} = 3$ . Note $n_{3} = 4 > 1$ and $n_{2} = 3 > 1$ , so Sylow alone does not show $S_{4}$ is non-simple — but $A_{4} ⊴ S_{4}$ does, by the proposition Subgroups of index two are normal.

□

Example 2.61 (A group of order 200 is not simple). Show that no group of order $200$ is simple.

Solution.

$200 = 2^{3} \cdot 5^{2}$ . Take $p = 5$ , so $p^{a} = 25$ and $m = 8$ .
By the theorem Sylow III: counting, $n_{5} ∣ 8$ and $n_{5} \equiv 1 (mod 5)$ . The divisors of $8$ are $1, 2, 4, 8$ , and of these only $1$ is congruent to $1$ modulo $5$ (since $2, 4, 8 \equiv 2, 4, 3$ ).
So $n_{5} = 1$ and the Sylow $5$ -subgroup is normal by the corollary A unique Sylow subgroup is normal.
It has order $25$ , so it is a proper nontrivial normal subgroup and $G$ is not simple.

□

Example 2.62 (Element counting rules out simplicity of order 30). Show that a group of order $30$ has a normal Sylow subgroup.

Solution.

$30 = 2 \cdot 3 \cdot 5$ . Sylow III allows $n_{3} \in {1, 10}$ (divisors of $10$ congruent to $1$ mod $3$ ) and $n_{5} \in {1, 6}$ (divisors of $6$ congruent to $1$ mod $5$ ).
Suppose both are greater than $1$ : $n_{3} = 10$ and $n_{5} = 6$ .
Distinct subgroups of order $3$ meet only in $e$ , so ten of them contribute $10 \cdot 2 = 20$ elements of order $3$ . Distinct subgroups of order $5$ likewise contribute $6 \cdot 4 = 24$ elements of order $5$ .
That is $20 + 24 + 1 = 45 > 30$ elements, which is impossible. So $n_{3} = 1$ or $n_{5} = 1$ , and the corresponding Sylow subgroup is normal.

□

Corollary 2.63 (Sylow recovers part of the converse to Lagrange). Let $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then for every $0 \leq b \leq a$ , $G$ has a subgroup of order $p^{b}$ .

Proof. By the theorem Sylow I: existence, $G$ has a subgroup $P$ of order $p^{a}$ . It therefore suffices to produce subgroups of every order $p^{b}$ inside the $p$ -group $P$ , which we do by induction on $a$ .

For $a = 0$ there is nothing to prove. Assume the claim for $p$ -groups of order less than $p^{a}$ . By the corollary Nontrivial $p$ -groups have nontrivial centre, $Z (P) \neq = {e}$ , so by the corollary The order of an element divides the order of the group it contains an element $z$ of order $p$ , and $⟨ z ⟩ ⊴ P$ by the proposition Every subgroup of an abelian group is normal; the centre is normal together with centrality. The quotient $P / ⟨ z ⟩$ has order $p^{a - 1}$ , so by induction it has a subgroup of order $p^{b - 1}$ for each $1 \leq b \leq a$ ; its preimage under the projection $π$ of the proposition Kernels are normal, and every normal subgroup is a kernel is a subgroup of $P$ of order $p \cdot p^{b - 1} = p^{b}$ . Together with the trivial subgroup for $b = 0$ , this gives all the required orders.∎

So the converse of Lagrange holds for prime-power divisors and fails for general divisors, with $A_{4}$ marking the boundary. That asymmetry is the honest summary of how much structure a finite group's order determines: prime powers are forced to appear, and everything else is a question that needs the group itself.

Exercises for §2.8 →

Common Mistakes

Forgetting to check nonemptiness. The empty set satisfies the closure conditions of the proposition Two-step subgroup test vacuously and is not a subgroup. Always exhibit an element, usually $e$ .
Assuming closure alone makes a subgroup. It does for finite subsets, by the proposition Finite subsets need only closure, and not otherwise: $N \subseteq Z$ is closed under $+$ and has no inverses.
Taking unions of subgroups. By the proposition Unions of subgroups are rarely subgroups, $H \cup K$ is a subgroup only when one contains the other; $2 Z \cup 3 Z$ misses $5$ . Use $⟨ H \cup K ⟩$ when you want a subgroup.
Treating a coset as a subgroup. Only $eH = H$ contains the identity; every other coset is a subset with no algebraic structure of its own. Cosets become elements of a group only after passing to $G / N$ for normal $N$ .
Reading Lagrange backwards. $∣ H ∣$ divides $∣ G ∣$ is a one-way implication. $A_{4}$ has order $12$ and no subgroup of order $6$ , so a divisor need not be realised; only prime powers are guaranteed, by the corollary Sylow recovers part of the converse to Lagrange.
Confusing normal with central. $g N g^{- 1} = N$ permits $g n g^{- 1} \neq = n$ . $A_{3} ⊴ S_{3}$ even though no transposition commutes with a $3$ -cycle.
Skipping the well-definedness check on $G / N$ . The formula $(a N) (b N) = ab N$ names representatives, and the example Failure of well-definedness for a non-normal subgroup shows two choices in $S_{3}$ giving different answers. The theorem The quotient group exists exactly when $N$ is normal is an "if and only if", and the "only if" direction is the one people assume away.
Believing $G / N$ is a subgroup of $G$ . Its elements are cosets, not elements of $G$ . There is a surjection $G \to G / N$ , never an inclusion in general — $Z /2$ is a quotient of $Z$ and is not a subgroup of it.
Using $[G : H] = ∣ G ∣/∣ H ∣$ for infinite groups. The quotient of two infinite cardinals is meaningless; $[Z : 4 Z] = 4$ must be computed by counting cosets, as in the theorem Tower law for indices.
Checking only one of the two Sylow conditions. $n_{p} \equiv 1 (mod p)$ alone leaves infinitely many candidates; $n_{p} ∣ m$ alone leaves all divisors. The pair is what usually pins $n_{p}$ to a single value.
Concluding a group is simple because Sylow fails to decide. In $S_{4}$ both $n_{2} = 3$ and $n_{3} = 4$ exceed $1$ , yet $A_{4}$ is a proper normal subgroup. Sylow gives sufficient conditions for non-simplicity, not necessary ones.
Assuming all Sylow $p$ -subgroups for different primes are conjugate. The theorem Sylow II: conjugacy and containment compares subgroups for the same prime only; a Sylow $2$ -subgroup and a Sylow $3$ -subgroup have different orders and cannot be conjugate.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter