Contents / Abstract Algebra / Group Homomorphisms and Isomorphisms

Chapter 3

Group Homomorphisms and Isomorphisms

Homomorphisms and what they preserve; the kernel proved normal and every normal subgroup realised as a kernel; injectivity equivalent to a trivial kernel; the first isomorphism theorem with its well-definedness spelled out, then the second, third and correspondence theorems; Cayley's theorem; and the automorphism group, with Inn(G) = G/Z(G) and Aut(Z/nZ) = (Z/nZ)^*.

Introduction

A group on its own is a static object: a set, a table, a list of orders. The subject only becomes dynamic once we can compare two groups, and the comparison has to respect the one thing a group actually is — its operation. That is what a homomorphism is. It is not a map that happens to be convenient; it is the only kind of map that carries any algebraic information from one group to another, and everything in this chapter is a consequence of that single requirement.

The chapter has a shape worth knowing in advance. We begin with the definition and squeeze out of it, for free, every fact that does not need a hypothesis: identity goes to identity, inverses to inverses, orders divide. Then we attach to each homomorphism its two canonical subgroups, the kernel and the image, and prove the two facts that make them worth naming — the image is a subgroup, and the kernel is normal. The second of these closes a loop opened in Subgroups, Cosets, and Quotient Groups, where normal subgroups were defined by a conjugation condition that looked arbitrary. It was not arbitrary: normal subgroups are exactly the kernels, and we shall prove both halves of that equivalence.

The centre of the chapter is the First Isomorphism Theorem, which says that a homomorphism is nothing but a quotient map followed by an inclusion. Every homomorphic image of $G$ is a quotient of $G$ , and every quotient of $G$ is a homomorphic image; the two words describe the same construction from opposite sides. From it the second and third isomorphism theorems and the correspondence theorem follow with almost no extra work, and those four results between them do most of the load-bearing in any structural argument about groups.

Two further themes close the chapter. Cayley's theorem shows that the abstract definition of a group bought us no new objects at all: every group is a group of permutations. And the automorphisms of a single group — its self-symmetries — form a group of their own, with the inner automorphisms sitting inside as a normal subgroup isomorphic to $G / Z (G)$ . The computational payoff is $Aut (Z / n Z) ≅ (Z / n Z)^{\times}$ , which turns a question about self-maps of a group into a count of units modulo $n$ .

Throughout, $G$ , $H$ and $K$ denote groups, written multiplicatively unless the notation says otherwise; $e_{G}$ is the identity of $G$ , abbreviated to $e$ when no confusion is possible; and $∣ g ∣$ denotes the order of an element.

3.1Homomorphisms and what they preserve for free

A group is a set together with an operation, so a map between groups deserves the name "map of groups" only if it does not disturb the operation. There is one way to write that down.

Definition 3.1 (Group homomorphism). Let $G$ and $H$ be groups. A function $ϕ : G \to H$ is a homomorphism if

ϕ (ab) = ϕ (a) ϕ (b) for all a, b \in G .

The product $ab$ on the left is taken in $G$ ; the product $ϕ (a) ϕ (b)$ on the right is taken in $H$ . The set of all homomorphisms $G \to H$ is written $Hom (G, H)$ .

The two sides of that equation live in different groups and are computed by different rules, which is precisely the point: the condition says the two rules agree under $ϕ$ . When the groups are written additively the same condition reads $ϕ (a + b) = ϕ (a) + ϕ (b)$ , and when one is additive and the other multiplicative it reads $ϕ (a + b) = ϕ (a) ϕ (b)$ — the exponential map is the standard example and it is a homomorphism for exactly this reason.

Definition 3.2 (Monomorphism, epimorphism, isomorphism, endomorphism). A homomorphism $ϕ : G \to H$ is a monomorphism (or an embedding) if it is injective, an epimorphism if it is surjective, and an isomorphism if it is bijective. If an isomorphism $G \to H$ exists we write $G ≅ H$ and say $G$ and $H$ are isomorphic. A homomorphism $G \to G$ is an endomorphism, and a bijective one is an automorphism.

Isomorphic groups are the same group written with different symbols. Any statement expressible purely in terms of the operation — the order of the group, the order of each element, abelianness, the number of subgroups of each order, simplicity — transfers across an isomorphism, because the isomorphism translates every product on one side into a product on the other. Anything not expressible that way, such as the literal names of the elements, does not transfer and is not part of the group's structure.

Proposition 3.3 (A homomorphism preserves the identity, inverses and powers). Let $ϕ : G \to H$ be a homomorphism. Then

$ϕ (e_{G}) = e_{H}$ ;
$ϕ (a^{- 1}) = ϕ (a)^{- 1}$ for every $a \in G$ ;
$ϕ (a^{n}) = ϕ (a)^{n}$ for every $a \in G$ and every $n \in Z$ .

Proof. (1) Apply the defining property to $a = b = e_{G}$ : $ϕ (e_{G}) = ϕ (e_{G} e_{G}) = ϕ (e_{G}) ϕ (e_{G})$ . This is an equation in the group $H$ , where $ϕ (e_{G})$ has an inverse; multiplying both sides on the left by $ϕ (e_{G})^{- 1}$ gives $e_{H} = ϕ (e_{G})$ .

(2) Fix $a \in G$ . Then $ϕ (a) ϕ (a^{- 1}) = ϕ (a a^{- 1}) = ϕ (e_{G}) = e_{H}$ by (1), and symmetrically $ϕ (a^{- 1}) ϕ (a) = e_{H}$ . An element of a group has exactly one inverse, so $ϕ (a^{- 1}) = ϕ (a)^{- 1}$ .

(3) For $n \geq 0$ induct: $n = 0$ is (1), and $ϕ (a^{n + 1}) = ϕ (a^{n} a) = ϕ (a^{n}) ϕ (a) = ϕ (a)^{n} ϕ (a) = ϕ (a)^{n + 1}$ . For $n < 0$ write $n = - m$ with $m > 0$ ; then $ϕ (a^{- m}) = ϕ ((a^{m})^{- 1}) = ϕ (a^{m})^{- 1} = (ϕ (a)^{m})^{- 1} = ϕ (a)^{- m}$ using (2) and the positive case.∎

This proposition is the reason the definition of a homomorphism contains only one clause. A structure-preserving map ought to be required to preserve the identity and the inverse operation as well, since both are part of the data of a group; the content of the proposition is that requiring them would be redundant. In practice this cuts the work of checking a candidate map by two thirds, and it supplies the fastest available disproof: if $ϕ (e_{G}) \neq = e_{H}$ , stop, the map is not a homomorphism.

Pitfall. The redundancy is special to groups. For a map of monoids the identity clause is genuinely independent and must be imposed by hand — the argument in part (1) cancelled $ϕ (e_{G})$ , which requires it to be invertible. The same caution applies to maps of rings, where $ϕ (1) = 1$ is an extra axiom, not a theorem.

Example 3.4 (The exponential map). Show that $exp : (R, +) \to (R^{> 0}, \times)$ , $x \mapsto e^{x}$ , is an isomorphism.

Solution.

The domain is additive and the codomain multiplicative, so the condition to check is $exp (x + y) = exp (x) exp (y)$ . The exponent law gives $e^{x + y} = e^{x} e^{y}$ for all real $x, y$ , so $exp$ is a homomorphism.
It is injective because $e^{x}$ is strictly increasing, and surjective onto $(0, \infty)$ because $e^{x}$ is continuous with $e^{x} \to 0$ as $x \to - \infty$ and $e^{x} \to \infty$ as $x \to \infty$ . So it is an isomorphism, with inverse $ln$ .
The free consequences check out, as Proposition A homomorphism preserves the identity, inverses and powers guarantees: $exp (0) = 1$ , the identity of $R^{> 0}$ , and $exp (- x) = 1/ e^{x} = exp (x)^{- 1}$ .

So the additive group of the reals and the multiplicative group of the positive reals are the same group. This is what a slide rule is: a physical realisation of the isomorphism, adding lengths in order to multiply numbers.□

Example 3.5 (Two maps that fail). Decide whether $f : (Z, +) \to (Z, +)$ , $f (x) = x + 1$ , and $g : (Z, +) \to (Z, +)$ , $g (x) = x^{2}$ , are homomorphisms.

Solution.

For $f$ , the fastest test is the identity: $f (0) = 1 \neq = 0$ , so by Proposition A homomorphism preserves the identity, inverses and powers $f$ is not a homomorphism. Directly: $f (x + y) = x + y + 1$ while $f (x) + f (y) = x + y + 2$ .
For $g$ the identity test is inconclusive, since $g (0) = 0$ . Test the operation: $g (x + y) = (x + y)^{2} = x^{2} + 2 x y + y^{2}$ , while $g (x) + g (y) = x^{2} + y^{2}$ . These agree only when $2 x y = 0$ , and $g (1 + 1) = 4 \neq = 2 = g (1) + g (1)$ .
Squaring is a homomorphism on any abelian group written multiplicatively, since $(ab)^{2} = a^{2} b^{2}$ there. The failure above is caused by mixing the additive structure of $Z$ with a multiplicative operation, not by squaring as such.

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Proposition 3.6 (Composites and inverses of homomorphisms). Let $ϕ : G \to H$ and $ψ : H \to K$ be homomorphisms. Then $ψ \circ ϕ : G \to K$ is a homomorphism. If $ϕ$ is an isomorphism, then $ϕ^{- 1} : H \to G$ is an isomorphism. Consequently $≅$ is an equivalence relation on any set of groups.

Proof. For the composite, $(ψ \circ ϕ) (ab) = ψ (ϕ (a) ϕ (b)) = ψ (ϕ (a)) ψ (ϕ (b))$ , using first that $ϕ$ is a homomorphism and then that $ψ$ is.

Suppose $ϕ$ is a bijective homomorphism and let $x, y \in H$ . Put $a = ϕ^{- 1} (x)$ and $b = ϕ^{- 1} (y)$ . Then $ϕ (ab) = ϕ (a) ϕ (b) = x y$ , so applying $ϕ^{- 1}$ gives $ϕ^{- 1} (x y) = ab = ϕ^{- 1} (x) ϕ^{- 1} (y)$ . Thus $ϕ^{- 1}$ is a homomorphism, and it is bijective, hence an isomorphism.

For the equivalence relation: $id_{G}$ is an isomorphism, giving reflexivity; the previous paragraph gives symmetry; and the composite of two isomorphisms is a bijective homomorphism, giving transitivity.∎

Proposition 3.7 (The order of an image divides the order of the element). Let $ϕ : G \to H$ be a homomorphism and let $a \in G$ have finite order $n$ . Then $ϕ (a)$ has finite order dividing $n$ . If $ϕ$ is injective, then $∣ ϕ (a) ∣ = ∣ a ∣$ .

Proof. By part (3) of Proposition A homomorphism preserves the identity, inverses and powers, $ϕ (a)^{n} = ϕ (a^{n}) = ϕ (e_{G}) = e_{H}$ , so $ϕ (a)$ has finite order, and its order divides $n$ because the set of integers $k$ with $ϕ (a)^{k} = e_{H}$ is the subgroup $m Z$ of $Z$ generated by $m = ∣ ϕ (a) ∣$ , which therefore contains $n$ .

If $ϕ$ is injective, let $m = ∣ ϕ (a) ∣$ . Then $ϕ (a^{m}) = ϕ (a)^{m} = e_{H} = ϕ (e_{G})$ , and injectivity forces $a^{m} = e_{G}$ , so $n ∣ m$ . With $m ∣ n$ this gives $m = n$ .∎

This little proposition does an enormous amount of work in practice. It is a constraint: before constructing a homomorphism you can often see that none exists, because some element's order cannot be accommodated. There is no non-trivial homomorphism $Z /3 Z \to Z /5 Z$ , because the image of a generator would have order dividing both $3$ and $5$ . It also tells you that isomorphic groups have the same number of elements of each order, which is the cheapest way to distinguish two groups of the same size.

Example 3.8 (Distinguishing two groups of order 4). Show that $Z /4 Z$ and the Klein four-group $V = Z /2 Z \times Z /2 Z$ are not isomorphic.

Solution.

Both have order $4$ , so counting alone does not separate them.
In $Z /4 Z$ the element $1$ has order $4$ . In $V$ every non-identity element $(1, 0), (0, 1), (1, 1)$ satisfies $x + x = (0, 0)$ , so every element has order $1$ or $2$ ; $V$ has no element of order $4$ .
An isomorphism is injective, so by Proposition The order of an image divides the order of the element it would carry the element of order $4$ to an element of order $4$ . There is none. Hence $Z /4 Z \neq ≅ V$ .

These are in fact the only two groups of order $4$ up to isomorphism, so the order sequence settles the classification in this size completely.□

Theorem 3.9 (Homomorphisms out of a cyclic group). Let $G = ⟨ a ⟩$ be cyclic and let $H$ be any group.

A homomorphism $ϕ : G \to H$ is determined by the single element $ϕ (a)$ .
If $G$ is infinite cyclic, then for every $h \in H$ there is exactly one homomorphism with $ϕ (a) = h$ ; hence $Hom (Z, H)$ is in bijection with $H$ .
If $∣ G ∣ = n$ , then $ϕ (a) = h$ defines a homomorphism if and only if $h^{n} = e_{H}$ .

Proof. (1) Every element of $G$ is $a^{k}$ for some integer $k$ , and $ϕ (a^{k}) = ϕ (a)^{k}$ by Proposition A homomorphism preserves the identity, inverses and powers. So the value of $ϕ$ everywhere is forced by $ϕ (a)$ .

(2) Given $h \in H$ , define $ϕ (a^{k}) = h^{k}$ . This is well defined because in an infinite cyclic group $a^{k} = a^{ℓ}$ forces $k = ℓ$ , so each element has exactly one exponent. It is a homomorphism since $ϕ (a^{k} a^{ℓ}) = h^{k + ℓ} = h^{k} h^{ℓ}$ . Uniqueness is (1).

(3) If $ϕ$ is a homomorphism with $ϕ (a) = h$ , then $h^{n} = ϕ (a^{n}) = ϕ (e) = e_{H}$ by Proposition The order of an image divides the order of the element. Conversely suppose $h^{n} = e_{H}$ and set $ϕ (a^{k}) = h^{k}$ . For well-definedness, suppose $a^{k} = a^{ℓ}$ . Then $a^{k - ℓ} = e$ , so $n ∣ k - ℓ$ , say $k - ℓ = q n$ ; hence $h^{k - ℓ} = (h^{n})^{q} = e_{H}$ and $h^{k} = h^{ℓ}$ . So the rule is unambiguous, and it preserves products exactly as in (2).∎

Corollary 3.10 (Counting homomorphisms between finite cyclic groups). For all $m, n \geq 1$ ,

Hom (Z / m Z, Z / n Z) = g cd (m, n) .

Proof. Write both groups additively, with $Z / m Z = ⟨ 1 ⟩$ . By Theorem Homomorphisms out of a cyclic group, the homomorphisms correspond exactly to the elements $h \in Z / n Z$ with $mh \equiv 0 (mod n)$ . That congruence says $n ∣ mh$ ; writing $d = g cd (m, n)$ and $m = d m^{'}$ , $n = d n^{'}$ with $g cd (m^{'}, n^{'}) = 1$ , it becomes $n^{'} ∣ m^{'} h$ , hence $n^{'} ∣ h$ . The multiples of $n^{'}$ in $Z / n Z$ are $0, n^{'}, 2 n^{'}, \dots, (d - 1) n^{'}$ , which is exactly $d$ elements, and each of them does satisfy $mh = d m^{'} \cdot k n^{'} = m^{'} k n \equiv 0$ . So the count is $d = g cd (m, n)$ .∎

Example 3.11 (Counting homomorphisms $Z /12 Z \to Z /18 Z$ ). How many homomorphisms are there from $Z /12 Z$ to $Z /18 Z$ , and which of them are injective?

Solution.

By Corollary Counting homomorphisms between finite cyclic groups the number is $g cd (12, 18) = 6$ .
Explicitly, $ϕ (1)$ must satisfy $12 ϕ (1) \equiv 0 (mod 18)$ , i.e. $18 ∣ 12 ϕ (1)$ , i.e. $3 ∣ 2 ϕ (1)$ , i.e. $3 ∣ ϕ (1)$ . The admissible values are $ϕ (1) \in {0, 3, 6, 9, 12, 15}$ — six of them, as predicted.
None is injective. An injective map would need $∣ ϕ (1) ∣ = 12$ by Proposition The order of an image divides the order of the element, but element orders in $Z /18 Z$ divide $18$ , and $12 ∤ 18$ . Indeed the largest order occurring among the six candidates is $∣3∣ = 6$ .

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Intuition. A homomorphism is a translation between two languages that is guaranteed to respect grammar: translate two words and combine them, or combine them and translate, and you get the same sentence. It need not be a dictionary — it may translate many words to one, and it may not reach every word in the target language. Those two defects are exactly the kernel and the image, and measuring them is the business of the next section.

Exercises for §3.1 →

3.2Kernel and image

Two subsets are attached to every homomorphism without any choice being made: the part of the domain that is destroyed, and the part of the codomain that is reached. Almost every question about a homomorphism reduces to a question about these two.

Definition 3.12 (Kernel and image). Let $ϕ : G \to H$ be a homomorphism. Its kernel is

ker ϕ = {g \in G : ϕ (g) = e_{H}} \subseteq G,

and its image is

im ϕ = ϕ (G) = {ϕ (g) : g \in G} \subseteq H .

Proposition 3.13 (The image is a subgroup; preimages are subgroups). Let $ϕ : G \to H$ be a homomorphism.

If $A \leq G$ then $ϕ (A) \leq H$ . In particular $im ϕ \leq H$ .
If $B \leq H$ then $ϕ^{- 1} (B) \leq G$ . If moreover $B ⊴ H$ , then $ϕ^{- 1} (B) ⊴ G$ .

Proof. (1) The set $ϕ (A)$ contains $ϕ (e_{G}) = e_{H}$ , so it is non-empty. If $x, y \in ϕ (A)$ , write $x = ϕ (a)$ , $y = ϕ (b)$ with $a, b \in A$ . Then $x y^{- 1} = ϕ (a) ϕ (b)^{- 1} = ϕ (a) ϕ (b^{- 1}) = ϕ (a b^{- 1})$ by Proposition A homomorphism preserves the identity, inverses and powers, and $a b^{- 1} \in A$ because $A$ is a subgroup. So $x y^{- 1} \in ϕ (A)$ , and by the subgroup criterion $ϕ (A) \leq H$ . Taking $A = G$ gives the image.

(2) Since $ϕ (e_{G}) = e_{H} \in B$ , the set $ϕ^{- 1} (B)$ contains $e_{G}$ . If $a, b \in ϕ^{- 1} (B)$ then $ϕ (a b^{- 1}) = ϕ (a) ϕ (b)^{- 1} \in B$ , because $B$ is a subgroup; so $a b^{- 1} \in ϕ^{- 1} (B)$ and $ϕ^{- 1} (B) \leq G$ .

Suppose now $B ⊴ H$ , let $a \in ϕ^{- 1} (B)$ and $g \in G$ . Then

ϕ (g a g^{- 1}) = ϕ (g) ϕ (a) ϕ (g)^{- 1} \in B

because $ϕ (a) \in B$ and $B$ is closed under conjugation by every element of $H$ , in particular by $ϕ (g)$ . Hence $g a g^{- 1} \in ϕ^{- 1} (B)$ , and $ϕ^{- 1} (B) ⊴ G$ .∎

The kernel is the special case $B = {e_{H}}$ , and the trivial subgroup is normal in $H$ , so part (2) already contains the next theorem. It is worth stating and proving separately anyway, because it is the result that explains where normality came from.

Theorem 3.14 (The kernel is a normal subgroup). Let $ϕ : G \to H$ be a homomorphism. Then $ker ϕ ⊴ G$ .

Proof. Subgroup. $ϕ (e_{G}) = e_{H}$ , so $e_{G} \in ker ϕ$ and the kernel is non-empty. If $a, b \in ker ϕ$ then

ϕ (a b^{- 1}) = ϕ (a) ϕ (b)^{- 1} = e_{H} e_{H}^{- 1} = e_{H},

so $a b^{- 1} \in ker ϕ$ . By the subgroup criterion, $ker ϕ \leq G$ .

Normal. Let $k \in ker ϕ$ and let $g \in G$ be arbitrary. Then

ϕ (g k g^{- 1}) = ϕ (g) ϕ (k) ϕ (g)^{- 1} = ϕ (g) e_{H} ϕ (g)^{- 1} = ϕ (g) ϕ (g)^{- 1} = e_{H} .

So $g k g^{- 1} \in ker ϕ$ for every $g \in G$ and every $k \in ker ϕ$ , that is, $g (ker ϕ) g^{- 1} \subseteq ker ϕ$ for all $g$ . Applying this with $g^{- 1}$ in place of $g$ gives $g^{- 1} (ker ϕ) g \subseteq ker ϕ$ , hence $ker ϕ \subseteq g (ker ϕ) g^{- 1}$ , and the two inclusions give equality. Therefore $ker ϕ ⊴ G$ .∎

Remark. The one-line calculation $ϕ (g k g^{- 1}) = ϕ (g) e_{H} ϕ (g)^{- 1} = e_{H}$ is the whole of it, and it explains the definition of normality rather than merely verifying it. Conjugation is the operation a homomorphism must respect because it is built out of multiplication; the identity is the one element a homomorphism cannot help but respect; so the set of elements hitting the identity cannot help but be conjugation-stable.

The converse is equally important, and together the two statements say that the notions "normal subgroup" and "kernel" have exactly the same extension.

Theorem 3.15 (Every normal subgroup is a kernel). Let $N ⊴ G$ . The canonical projection

π : G \to G / N, π (g) = g N,

is a surjective homomorphism with $ker π = N$ . Hence a subgroup of $G$ is normal if and only if it is the kernel of some homomorphism defined on $G$ .

Proof. The quotient $G / N$ is a group precisely because $N$ is normal, with the operation $(a N) (b N) = ab N$ , as shown in Subgroups, Cosets, and Quotient Groups. That formula is exactly the statement $π (ab) = π (a) π (b)$ , so $π$ is a homomorphism, and it is surjective because every coset $g N$ is $π (g)$ .

Its kernel: the identity of $G / N$ is the coset $N = e N$ , so

ker π = {g \in G : g N = N} = {g \in G : g \in N} = N,

using the standard fact that $g N = N$ if and only if $g \in N$ .

For the final sentence, Theorem The kernel is a normal subgroup gives one direction and the projection $π$ gives the other.∎

Theorem 3.16 (Injectivity is detected by the kernel). A homomorphism $ϕ : G \to H$ is injective if and only if $ker ϕ = {e_{G}}$ .

Proof. $(\Rightarrow)$ Suppose $ϕ$ is injective and let $k \in ker ϕ$ . Then $ϕ (k) = e_{H} = ϕ (e_{G})$ , the second equality by Proposition A homomorphism preserves the identity, inverses and powers. Injectivity applied to these two elements with the same image gives $k = e_{G}$ . So $ker ϕ \subseteq {e_{G}}$ , and the reverse inclusion always holds, whence $ker ϕ = {e_{G}}$ .

$(\Leftarrow)$ Suppose $ker ϕ = {e_{G}}$ and let $a, b \in G$ satisfy $ϕ (a) = ϕ (b)$ . Multiply both sides on the right by $ϕ (b)^{- 1}$ :

ϕ (a) ϕ (b)^{- 1} = e_{H} .

The left-hand side is $ϕ (a) ϕ (b^{- 1}) = ϕ (a b^{- 1})$ , so $a b^{- 1} \in ker ϕ = {e_{G}}$ , giving $a b^{- 1} = e_{G}$ and therefore $a = b$ . Hence $ϕ$ is injective.∎

This is the workhorse of the chapter, and it is worth seeing why it is not a triviality. Injectivity is a statement about all pairs of elements: there are $∣ G ∣^{2}$ of them to inspect. The kernel condition is a statement about a single element, the identity, and its preimage. The theorem collapses a quadratic amount of checking to a linear amount, and it does so because the group operation lets any pair $(a, b)$ be translated to the pair $(a b^{- 1}, e)$ without loss. No such reduction exists for arbitrary functions between sets; it is a dividend paid by the group structure. The same argument, in the same three lines, is what makes "a linear map is injective iff its null space is zero" true in linear algebra, as in Linear Transformations.

Proposition 3.17 (The fibres of a homomorphism are the cosets of the kernel). Let $ϕ : G \to H$ be a homomorphism with $K = ker ϕ$ , and let $a \in G$ . Then

ϕ^{- 1} ({ϕ (a)}) = a K = K a .

In particular all non-empty fibres of $ϕ$ have the same cardinality, namely $∣ K ∣$ .

Proof. If $x = ak$ with $k \in K$ , then $ϕ (x) = ϕ (a) ϕ (k) = ϕ (a) e_{H} = ϕ (a)$ , so $a K \subseteq ϕ^{- 1} ({ϕ (a)})$ . Conversely, if $ϕ (x) = ϕ (a)$ then, as in Theorem Injectivity is detected by the kernel, $ϕ (a^{- 1} x) = ϕ (a)^{- 1} ϕ (x) = e_{H}$ , so $k := a^{- 1} x \in K$ and $x = ak \in a K$ . Thus the fibre is exactly $a K$ , and $a K = K a$ since $K$ is normal by Theorem The kernel is a normal subgroup.

The map $k \mapsto ak$ is a bijection $K \to a K$ , with inverse $x \mapsto a^{- 1} x$ , so $∣ a K ∣ = ∣ K ∣$ .∎

Figure 3.18. Every fibre of

ϕ

is a coset of

ker ϕ

, and the whole coset collapses onto a single point of

im ϕ

. The fibres partition

G

into blocks of equal size

∣ ker ϕ ∣

Corollary 3.19 (The counting formula). Let $ϕ : G \to H$ be a homomorphism with $G$ finite. Then

∣ G ∣ = ∣ ker ϕ ∣ \cdot ∣ im ϕ ∣.

Consequently $∣ im ϕ ∣$ divides both $∣ G ∣$ and, when $H$ is finite, $∣ H ∣$ .

Proof. By Proposition The fibres of a homomorphism are the cosets of the kernel, the sets $ϕ^{- 1} ({y})$ for $y \in im ϕ$ partition $G$ , and each has exactly $∣ ker ϕ ∣$ elements. There are $∣ im ϕ ∣$ of them, so $∣ G ∣ = ∣ im ϕ ∣ \cdot ∣ ker ϕ ∣$ .

The image is a subgroup of $H$ by Proposition The image is a subgroup; preimages are subgroups, so Lagrange's theorem, proved in Subgroups, Cosets, and Quotient Groups, gives $∣ im ϕ ∣ ∣ ∣ H ∣$ .∎

This is the single most useful computational tool in the chapter. It turns a structural question into arithmetic, and combined with divisibility it frequently forces a homomorphism to be trivial before any calculation is attempted: if $∣ G ∣ = 7$ and $∣ H ∣ = 12$ , the image has order dividing $g cd (7, 12) = 1$ , so the only homomorphism $G \to H$ is the constant map.

Example 3.20 (Kernel and image of the determinant). Compute the kernel and image of $det : G L_{2} (R) \to R^{\times}$ , and say what the first isomorphism theorem will give.

Solution.

It is a homomorphism because $det (A B) = det (A) det (B)$ , which is the multiplicativity of the determinant proved in Determinants.
The identity of $R^{\times}$ is $1$ , so $ker (det) = {A : det A = 1} = S L_{2} (R)$ , the special linear group.
Every $r \neq = 0$ occurs: $det (r 0 01) = r$ . So $im (det) = R^{\times}$ and $det$ is surjective.
Because it is a kernel, $S L_{2} (R) ⊴ G L_{2} (R)$ automatically, by Theorem The kernel is a normal subgroup — no conjugation check needed. The First Isomorphism Theorem will give $G L_{2} (R) / S L_{2} (R) ≅ R^{\times}$ .

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Example 3.21 (Kernel and image of the sign homomorphism). Let $sgn : S_{4} \to {\pm 1}$ send a permutation to $+ 1$ if it is even and $- 1$ if it is odd. Find $∣ ker sgn ∣$ .

Solution.

$sgn$ is a homomorphism and is surjective, since the transposition $(12)$ has sign $- 1$ and the identity has sign $+ 1$ . So $∣ im sgn ∣ = 2$ .
By Corollary The counting formula, $∣ ker sgn ∣ = ∣ S_{4} ∣/2 = 24/2 = 12$ .
The kernel is by definition the set of even permutations, that is the alternating group $A_{4}$ , and we have just computed $∣ A_{4} ∣ = 12$ without listing a single permutation.

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Example 3.22 (A homomorphism forced to be trivial). Show that the only homomorphism $ϕ : Z /7 Z \to S_{3}$ is the trivial one.

Solution.

$im ϕ$ is a subgroup of $S_{3}$ , so $∣ im ϕ ∣$ divides $∣ S_{3} ∣ = 6$ by Lagrange.
By Corollary The counting formula, $∣ im ϕ ∣$ also divides $∣ Z /7 Z ∣ = 7$ .
So $∣ im ϕ ∣$ divides $g cd (6, 7) = 1$ , forcing $∣ im ϕ ∣ = 1$ : every element is sent to the identity.

The same argument shows there is no non-trivial homomorphism between groups of coprime order in either direction.□

Example 3.23 (Reading off injectivity and surjectivity). Let $ϕ : Z \to Z /8 Z$ be $ϕ (k) = 2 k mod 8$ . Find $ker ϕ$ and $im ϕ$ , and decide whether $ϕ$ is injective or surjective.

Solution.

It is a homomorphism: $ϕ (k + ℓ) = 2 (k + ℓ) = 2 k + 2 ℓ = ϕ (k) + ϕ (ℓ)$ in $Z /8 Z$ .
Kernel: $2 k \equiv 0 (mod 8) ⟺ 8 ∣ 2 k ⟺ 4 ∣ k$ , so $ker ϕ = 4 Z$ . This is not ${0}$ , so by Theorem Injectivity is detected by the kernel $ϕ$ is not injective.
Image: as $k$ runs over $Z$ , $2 k mod 8$ takes the values ${0, 2, 4, 6}$ , a subgroup of order $4$ . Since $4 \neq = 8$ , $ϕ$ is not surjective.
The counting formula does not apply here ( $G$ is infinite), but the First Isomorphism Theorem will: $Z /4 Z ≅ {0, 2, 4, 6}$ , and both sides have order $4$ .

□

Intuition. Think of $ϕ$ as a lossy photocopier. The kernel is everything that comes out blank — it measures what is thrown away, and a trivial kernel means nothing is lost, so the copy is faithful. The image is the set of pages that can actually be produced — what is reached. Proposition The fibres of a homomorphism are the cosets of the kernel says the loss is perfectly uniform: every output page comes from exactly $∣ ker ϕ ∣$ input pages, never more and never fewer. That uniformity is what makes the counting formula a product rather than an inequality.

Exercises for §3.2 →

3.3The isomorphism theorems

We have two facts pointing at each other: every kernel is normal, and every normal subgroup is a kernel. The First Isomorphism Theorem makes the correspondence precise by identifying the quotient with the image, and it does so in a way that is constructive — it names the isomorphism rather than merely asserting one exists.

Theorem 3.24 (First isomorphism theorem). Let $ϕ : G \to H$ be a homomorphism and put $K = ker ϕ$ . Then the map

\overline{ϕ} : G / K ⟶ im ϕ, \overline{ϕ} (g K) = ϕ (g),

is a well-defined isomorphism. Hence

G / ker ϕ ≅ im ϕ,

and $ϕ = ι \circ \overline{ϕ} \circ π$ , where $π : G \to G / K$ is the canonical projection and $ι : im ϕ ↪ H$ the inclusion.

Proof. Well-definedness — the crux. The formula $\overline{ϕ} (g K) = ϕ (g)$ defines $\overline{ϕ}$ on a coset by choosing a representative $g$ of that coset, and a coset has many representatives. Nothing yet guarantees that two representatives give the same answer, and if they did not, $\overline{ϕ}$ would not be a function at all. So suppose $g K = g^{'} K$ ; we must show $ϕ (g) = ϕ (g^{'})$ . From $g K = g^{'} K$ we get $g^{- 1} g^{'} \in K = ker ϕ$ , hence

e_{H} = ϕ (g^{- 1} g^{'}) = ϕ (g)^{- 1} ϕ (g^{'}),

and multiplying on the left by $ϕ (g)$ gives $ϕ (g) = ϕ (g^{'})$ . So the value does not depend on the representative, and $\overline{ϕ}$ is a well-defined function on $G / K$ . Note where the hypothesis was used: the argument works precisely because the subgroup we quotiented by is the kernel — for any smaller normal subgroup the step $ϕ (g^{- 1} g^{'}) = e_{H}$ would fail.

Homomorphism. Using the quotient operation $(a K) (b K) = ab K$ ,

\overline{ϕ} ((a K) (b K)) = \overline{ϕ} (ab K) = ϕ (ab) = ϕ (a) ϕ (b) = \overline{ϕ} (a K) \overline{ϕ} (b K) .

Surjectivity onto the image. Every element of $im ϕ$ is $ϕ (g)$ for some $g \in G$ , and $ϕ (g) = \overline{ϕ} (g K)$ .

Injectivity. By Theorem Injectivity is detected by the kernel it suffices to show $ker \overline{ϕ}$ is trivial. Suppose $\overline{ϕ} (g K) = e_{H}$ . Then $ϕ (g) = e_{H}$ , so $g \in K$ , so $g K = K$ , which is the identity element of $G / K$ . Hence $ker \overline{ϕ} = {K}$ and $\overline{ϕ}$ is injective.

Therefore $\overline{ϕ}$ is a bijective homomorphism, i.e. an isomorphism. The factorisation $ϕ = ι \circ \overline{ϕ} \circ π$ is immediate: $(ι \circ \overline{ϕ} \circ π) (g) = \overline{ϕ} (g K) = ϕ (g)$ .∎

Figure 3.25. The First Isomorphism Theorem as a commutative triangle: every homomorphism factors as the canonical projection followed by an injection, and the induced map

\overline{ϕ}

is an isomorphism onto the image.

The theorem is usually read as a computational device — "to identify a quotient, find a homomorphism whose kernel it is" — and that is how it will be used below. But its structural content is larger. It says that the two apparently different constructions of new groups from old, quotienting and taking images, produce exactly the same class of groups. There are no homomorphic images beyond the quotients, and no quotients beyond the images. The construction of $G / N$ , which looks like an artificial device for collapsing a subgroup, is thereby revealed as the universal target of all maps out of $G$ that kill $N$ .

Corollary 3.26 (The universal property of the quotient). Let $N ⊴ G$ and let $ϕ : G \to H$ be a homomorphism with $N \subseteq ker ϕ$ . Then there is a unique homomorphism $\overline{ϕ} : G / N \to H$ with $\overline{ϕ} \circ π = ϕ$ .

Proof. Existence. Define $\overline{ϕ} (g N) = ϕ (g)$ . If $g N = g^{'} N$ then $g^{- 1} g^{'} \in N \subseteq ker ϕ$ , so $ϕ (g)^{- 1} ϕ (g^{'}) = ϕ (g^{- 1} g^{'}) = e_{H}$ and $ϕ (g) = ϕ (g^{'})$ ; the map is well defined. It is a homomorphism by the computation in the proof of Theorem First isomorphism theorem, and $\overline{ϕ} (π (g)) = ϕ (g)$ by construction.

Uniqueness. If $ψ \circ π = ϕ$ then $ψ (g N) = ψ (π (g)) = ϕ (g)$ for every $g$ , and since $π$ is surjective this determines $ψ$ on all of $G / N$ .∎

Example 3.27 (Identifying $Z / n Z$ as a quotient). Use the First Isomorphism Theorem to prove that $Z / n Z$ is isomorphic to the group $μ_{n}$ of complex $n$ -th roots of unity under multiplication.

Solution.

Define $ϕ : Z \to C^{\times}$ by $ϕ (k) = e^{2 πik / n}$ . It is a homomorphism because $ϕ (k + ℓ) = e^{2 πi (k + ℓ) / n} = ϕ (k) ϕ (ℓ)$ .
Image: the values $ϕ (0), \dots, ϕ (n - 1)$ are the $n$ distinct $n$ -th roots of unity, and $ϕ (k)$ depends only on $k mod n$ , so $im ϕ = μ_{n}$ .
Kernel: $e^{2 πik / n} = 1$ exactly when $k / n \in Z$ , i.e. when $n ∣ k$ . So $ker ϕ = n Z$ .
By Theorem First isomorphism theorem, $Z / n Z = Z / ker ϕ ≅ im ϕ = μ_{n}$ .

The point is that we never had to construct the isomorphism by hand or check $n$ cases; writing down one homomorphism and computing its kernel did all of it.□

Example 3.28 (The quotient of $R$ by $Z$ is the circle). Identify $R / Z$ , where $R$ is additive.

Solution.

Let $ϕ : R \to C^{\times}$ be $ϕ (x) = e^{2 πi x}$ ; then $ϕ (x + y) = ϕ (x) ϕ (y)$ , so it is a homomorphism.
Its image is the unit circle $S^{1} = {z \in C : ∣ z ∣ = 1}$ , since every such $z$ is $e^{i θ}$ for some real $θ$ and $θ = 2 π x$ is solvable.
Its kernel is ${x : e^{2 πi x} = 1} = Z$ .
Hence $R / Z ≅ S^{1}$ : the real line with the integers collapsed is the circle group. Winding the line onto the circle is the projection $π$ , and the theorem says the winding map loses exactly the integer part.

□

Example 3.29 (A quotient of $S_{4}$ ). What is $S_{4} / A_{4}$ ?

Solution.

Use $sgn : S_{4} \to {\pm 1}$ , which is surjective with kernel $A_{4}$ , as in Example Kernel and image of the sign homomorphism.
Theorem First isomorphism theorem gives $S_{4} / A_{4} ≅ {\pm 1} ≅ Z /2 Z$ .
Checking with Corollary The counting formula: $∣ S_{4} / A_{4} ∣ = 24/12 = 2$ . ✓

□

Theorem 3.30 (Second isomorphism theorem (the diamond)). Let $G$ be a group, $H \leq G$ and $N ⊴ G$ . Then

$H N = {hn : h \in H, n \in N}$ is a subgroup of $G$ ;
$N ⊴ H N$ and $H \cap N ⊴ H$ ;
$H N / N ≅ H / (H \cap N)$ .

Proof. (1) $e = ee \in H N$ . Given $h_{1} n_{1}, h_{2} n_{2} \in H N$ ,

(h_{1} n_{1}) (h_{2} n_{2})^{- 1} = h_{1} n_{1} n_{2}^{- 1} h_{2}^{- 1} = (h_{1} h_{2}^{- 1}) (h_{2} n_{1} n_{2}^{- 1} h_{2}^{- 1}),

where the second bracket lies in $N$ because $N$ is normal in $G$ and $h_{2} \in G$ . So the product lies in $H N$ and $H N \leq G$ .

(2) $N \subseteq H N$ and $N$ is normal in $G$ , hence normal in the subgroup $H N$ . Normality of $H \cap N$ in $H$ will follow from (3), where it appears as a kernel.

(3) Define $θ : H \to H N / N$ by $θ (h) = h N$ ; this is the restriction to $H$ of the canonical projection $H N \to H N / N$ , hence a homomorphism.

Surjectivity. A typical element of $H N / N$ is $(hn) N$ with $h \in H$ , $n \in N$ . Since $n \in N$ we have $n N = N$ , so $(hn) N = h (n N) = h N = θ (h)$ . Every element of the quotient is hit.

Kernel. $θ (h) = N$ if and only if $h N = N$ , i.e. $h \in N$ . Combined with $h \in H$ this says $ker θ = H \cap N$ . By Theorem The kernel is a normal subgroup, $H \cap N ⊴ H$ , completing (2).

Now Theorem First isomorphism theorem applied to $θ$ gives $H / (H \cap N) ≅ im θ = H N / N$ .∎

The name "diamond" comes from the picture: $H N$ at the top, $H$ and $N$ below it, $H \cap N$ at the bottom. The theorem says the two opposite edges of this diamond — from $H \cap N$ up to $H$ , and from $N$ up to $H N$ — are the same quotient, not merely quotients of the same size. In the finite case it yields the order formula

∣ H N ∣ = \frac{∣ H ∣ ∣ N ∣}{∣ H \cap N ∣},

which is often the quickest way to show two subgroups cannot both sit inside a group of a given size.

Theorem 3.31 (Third isomorphism theorem (cancellation)). Let $N ⊴ G$ and $M ⊴ G$ with $N \subseteq M$ . Then $M / N ⊴ G / N$ and

(G / N) / (M / N) ≅ G / M .

Proof. Consider $ϕ : G / N \to G / M$ defined by $ϕ (g N) = g M$ .

Well defined. If $g N = g^{'} N$ then $g^{- 1} g^{'} \in N \subseteq M$ , hence $g M = g^{'} M$ . (This is the only place the hypothesis $N \subseteq M$ is used, and it is indispensable.)

Homomorphism. $ϕ ((a N) (b N)) = ϕ (ab N) = ab M = (a M) (b M) = ϕ (a N) ϕ (b N)$ .

Surjective. Every coset $g M$ is $ϕ (g N)$ .

Kernel. $ϕ (g N) = M$ , the identity of $G / M$ , if and only if $g \in M$ ; so

ker ϕ = {g N : g \in M} = M / N .

In particular $M / N$ is a kernel, hence normal in $G / N$ by Theorem The kernel is a normal subgroup. Applying Theorem First isomorphism theorem to $ϕ$ gives $(G / N) / (M / N) ≅ G / M$ .∎

Intuition. The third theorem is the schoolroom rule $\frac{a / c}{b / c} = \frac{a}{b}$ transplanted to groups: quotient by $N$ , then quotient again by $M / N$ , and the $N$ s cancel. The mechanism is the same in both settings — dividing twice by nested things is dividing once by the larger — and the proof above is just that sentence made precise by exhibiting the map that does the cancelling.

Example 3.32 (A concrete diamond in $Z /12 Z$ ). In $G = Z /12 Z$ take $H = ⟨ 3 ⟩ = {0, 3, 6, 9}$ and $N = ⟨ 4 ⟩ = {0, 4, 8}$ . Verify the second isomorphism theorem.

Solution.

$G$ is abelian, so every subgroup is normal and $H N = H + N$ is the subgroup generated by $3$ and $4$ . Since $g cd (3, 4) = 1$ , it contains $1$ , so $H N = G$ , of order $12$ .
$H \cap N$ consists of the elements that are multiples of both $3$ and $4$ modulo $12$ , i.e. multiples of $lcm (3, 4) = 12$ , so $H \cap N = {0}$ .
Left side: $H N / N = G / N$ has order $12/3 = 4$ . Right side: $H / (H \cap N) ≅ H$ has order $4$ . They agree, and both are cyclic of order $4$ .
The order formula also checks out: $∣ H N ∣ = ∣ H ∣∣ N ∣/∣ H \cap N ∣ = 4 \cdot 3/1 = 12 = ∣ G ∣$ . ✓

□

Example 3.33 (Cancelling in $Z /12 Z$ ). With $G = Z /12 Z$ , $N = ⟨ 6 ⟩ = {0, 6}$ and $M = ⟨ 2 ⟩ = {0, 2, 4, 6, 8, 10}$ , verify the third isomorphism theorem and give the order of $(G / N) / (M / N)$ .

Solution.

$N \subseteq M$ since $6 = 3 \cdot 2$ , and all subgroups are normal as $G$ is abelian.
$∣ N ∣ = 2$ , $∣ M ∣ = 6$ , so $∣ G / N ∣ = 6$ and $∣ M / N ∣ = 3$ .
Theorem Third isomorphism theorem gives $(G / N) / (M / N) ≅ G / M$ , of order $12/6 = 2$ .
Directly: $∣ (G / N) / (M / N) ∣ = 6/3 = 2$ . ✓ Both sides are $Z /2 Z$ .

□

Example 3.34 (An order that forces a contradiction). A group $G$ has order $30$ . Show that it cannot contain subgroups $H$ and $N$ with $∣ H ∣ = 10$ , $∣ N ∣ = 6$ and $H \cap N = {e}$ .

Solution.

If $N$ were normal, Theorem Second isomorphism theorem (the diamond) would give $∣ H N ∣ = ∣ H ∣∣ N ∣/∣ H \cap N ∣ = 10 \cdot 6/1 = 60$ .
But $H N \subseteq G$ and $∣ G ∣ = 30 < 60$ , a contradiction.
Even without normality the set $H N$ has $∣ H ∣∣ N ∣/∣ H \cap N ∣ = 60$ distinct elements by the same coset-counting argument, so the conclusion stands regardless: $H \cap N$ must have order at least $2$ .

□

Exercises for §3.3 →

3.4The correspondence theorem

The three isomorphism theorems describe individual quotients. The correspondence theorem describes the whole subgroup lattice of a quotient at once, and it is what makes quotients tractable: you never have to hunt for the subgroups of $G / N$ , because you already know the subgroups of $G$ that contain $N$ .

Theorem 3.35 (Correspondence (lattice isomorphism) theorem). Let $N ⊴ G$ with canonical projection $π : G \to G / N$ . The assignments

A ⟼ A / N = π (A), \overline{B} ⟼ π^{- 1} (\overline{B})

are mutually inverse, inclusion-preserving bijections between

{A : N \leq A \leq G} and {\overline{B} : \overline{B} \leq G / N} .

Moreover, under this bijection:

$A_{1} \leq A_{2}$ if and only if $A_{1} / N \leq A_{2} / N$ , and then $[A_{2} : A_{1}] = [A_{2} / N : A_{1} / N]$ ;
$A ⊴ G$ if and only if $A / N ⊴ G / N$ ;
$π (A_{1} \cap A_{2}) = π (A_{1}) \cap π (A_{2})$ and $π (⟨ A_{1}, A_{2} ⟩) = ⟨ π (A_{1}), π (A_{2})⟩$ .

Proof. The maps land where claimed. If $N \leq A \leq G$ then $π (A) \leq G / N$ by Proposition The image is a subgroup; preimages are subgroups. If $\overline{B} \leq G / N$ then $π^{- 1} (\overline{B}) \leq G$ by the same proposition, and it contains $π^{- 1} ({N}) = ker π = N$ .

$π^{- 1} (π (A)) = A$ for $N \leq A$ . The inclusion $\supseteq$ is trivial. Conversely let $g \in π^{- 1} (π (A))$ , so $g N = a N$ for some $a \in A$ . Then $a^{- 1} g \in N \subseteq A$ , hence $g = a (a^{- 1} g) \in A$ . This is the step where $N \leq A$ is essential; without it the operation $A \mapsto π^{- 1} (π (A))$ enlarges $A$ to $A N$ .

$π (π^{- 1} (\overline{B})) = \overline{B}$ . The inclusion $\subseteq$ is trivial, and $\supseteq$ holds because $π$ is surjective: given $\overline{b} \in \overline{B}$ , pick $g$ with $π (g) = \overline{b}$ ; then $g \in π^{- 1} (\overline{B})$ and $\overline{b} = π (g) \in π (π^{- 1} (\overline{B}))$ .

So the two maps are mutually inverse bijections, and both visibly preserve inclusion.

(1) Inclusion-preservation in both directions is immediate from the bijection. For the index, the cosets of $A_{1}$ in $A_{2}$ correspond to the cosets of $A_{1} / N$ in $A_{2} / N$ via $a A_{1} \mapsto π (a) (A_{1} / N)$ ; this is well defined and injective because $π (a) (A_{1} / N) = π (a^{'}) (A_{1} / N)$ means $a^{- 1} a^{'} \in π^{- 1} (A_{1} / N) = A_{1}$ , and it is surjective since $π$ is.

(2) If $A ⊴ G$ then for $g \in G$ , $π (g) π (A) π (g)^{- 1} = π (g A g^{- 1}) = π (A)$ , and every element of $G / N$ is some $π (g)$ , so $A / N ⊴ G / N$ . Conversely if $A / N ⊴ G / N$ then $A = π^{- 1} (A / N)$ is normal in $G$ by the last clause of Proposition The image is a subgroup; preimages are subgroups.

(3) Both follow from the bijection: $π^{- 1}$ preserves intersections of sets, and applying the bijection to $A_{1} \cap A_{2}$ , which is the largest subgroup contained in both, gives the largest subgroup contained in both images. The join statement is the dual argument.∎

Figure 3.36. The correspondence theorem: the subgroup lattice of

G / N

is a faithful copy of the part of the subgroup lattice of

G

lying above

N

. Indices along corresponding edges agree.

Corollary 3.37 (Simplicity of a quotient means maximality of the subgroup). Let $N ⊴ G$ . Then $G / N$ has no normal subgroups other than the trivial one and itself if and only if $N$ is a maximal normal subgroup of $G$ , that is, $N \neq = G$ and there is no $M ⊴ G$ with $N ⊊ M ⊊ G$ .

Proof. By part (2) of Theorem Correspondence (lattice isomorphism) theorem, the normal subgroups of $G / N$ are exactly the $M / N$ for $N \leq M ⊴ G$ , and by the bijection this correspondence is strict: $M / N$ is trivial iff $M = N$ , and equals $G / N$ iff $M = G$ . So $G / N$ has only the two trivial normal subgroups precisely when no intermediate $M$ exists.∎

Example 3.38 (The subgroups of $Z /12 Z$ from those of $Z$ ). Use the correspondence theorem to list the subgroups of $Z /12 Z$ .

Solution.

$Z /12 Z = Z / N$ with $N = 12 Z$ . The subgroups of $Z$ are exactly $d Z$ for $d \geq 0$ , and $12 Z \subseteq d Z$ if and only if $d ∣ 12$ .
So the subgroups above $N$ are $d Z$ for $d \in {1, 2, 3, 4, 6, 12}$ — six of them.
By Theorem Correspondence (lattice isomorphism) theorem the subgroups of $Z /12 Z$ are exactly $d Z /12 Z = ⟨ d ⟩$ for those six $d$ , of orders $12, 6, 4, 3, 2, 1$ respectively.
Inclusions correspond too: $⟨ 4 ⟩ \leq ⟨ 2 ⟩$ because $4 Z \subseteq 2 Z$ . The lattice of divisors of $12$ is the subgroup lattice.

□

Example 3.39 (Counting subgroups of a quotient). Let $G = Z /12 Z$ and $N = ⟨ 4 ⟩ = {0, 4, 8}$ . Determine the number of subgroups of $G / N$ and identify $G / N$ .

Solution.

$∣ N ∣ = 3$ , so $∣ G / N ∣ = 12/3 = 4$ .
By Theorem Correspondence (lattice isomorphism) theorem, the subgroups of $G / N$ are in bijection with the subgroups of $G$ containing $N$ . The subgroups of $G$ are $⟨ d ⟩$ for $d ∣ 12$ , and $⟨ d ⟩ \supseteq ⟨ 4 ⟩$ requires $4 \in ⟨ d ⟩$ , i.e. $d ∣ 4$ : so $d \in {1, 2, 4}$ , giving $⟨ 1 ⟩ = G$ , $⟨ 2 ⟩$ and $⟨ 4 ⟩ = N$ — three subgroups.
Hence $G / N$ has exactly $3$ subgroups. Of the two groups of order $4$ , the cyclic one has $3$ subgroups (orders $1, 2, 4$ ) and the Klein four-group has $5$ . So $G / N ≅ Z /4 Z$ .
The correspondence identified the isomorphism type of the quotient without ever multiplying two cosets together.

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Exercises for §3.4 →

3.5Cayley's theorem: every group is a permutation group

The axioms for a group were abstracted from the behaviour of permutations, and it is fair to ask whether the abstraction gained anything — whether there exist groups that are not, in disguise, groups of permutations. Cayley's theorem answers no. The proof is an application of the injectivity criterion and is barely a page.

Definition 3.40 (Left translation). For $g$ in a group $G$ , the left translation by $g$ is the map $λ_{g} : G \to G$ given by $λ_{g} (x) = gx$ .

Lemma 3.41 (Left translations are permutations). For every $g \in G$ the map $λ_{g}$ is a bijection from $G$ to itself, and $λ_{g} \circ λ_{h} = λ_{g h}$ for all $g, h \in G$ .

Proof. The map $λ_{g^{- 1}}$ is a two-sided inverse for $λ_{g}$ : $λ_{g^{- 1}} (λ_{g} (x)) = g^{- 1} (gx) = x$ and $λ_{g} (λ_{g^{- 1}} (x)) = g (g^{- 1} x) = x$ . So $λ_{g}$ is a bijection, i.e. an element of the symmetric group $Sym (G)$ of all permutations of the set $G$ .

For the second claim, $(λ_{g} \circ λ_{h}) (x) = λ_{g} (h x) = g (h x) = (g h) x = λ_{g h} (x)$ by associativity, for every $x$ .∎

Note that $λ_{g}$ is not a homomorphism $G \to G$ unless $g = e$ — it moves the identity to $g$ . It is a permutation of the underlying set, which is all that is needed.

Theorem 3.42 (Cayley's theorem). Every group $G$ embeds in the symmetric group on its own underlying set: the map

Λ : G \to Sym (G), Λ (g) = λ_{g},

is an injective homomorphism. Hence $G ≅ Λ (G) \leq Sym (G)$ .

Proof. Homomorphism. $Λ (g h) = λ_{g h} = λ_{g} \circ λ_{h} = Λ (g) Λ (h)$ by Lemma Left translations are permutations, the product in $Sym (G)$ being composition.

Injective. By Theorem Injectivity is detected by the kernel it is enough to compute the kernel. Suppose $Λ (g) = id$ , i.e. $gx = x$ for every $x \in G$ . Taking $x = e$ gives $g = e$ . So $ker Λ = {e}$ and $Λ$ is injective.

Conclusion. $Λ$ is injective, so it is an isomorphism onto its image, and its image is a subgroup of $Sym (G)$ by Proposition The image is a subgroup; preimages are subgroups. Hence $G ≅ Λ (G)$ .∎

Corollary 3.43 (Finite groups embed in $S_{n}$ ). If $∣ G ∣ = n$ then $G$ is isomorphic to a subgroup of $S_{n}$ .

Proof. Label the $n$ elements of $G$ as $1, \dots, n$ ; this identifies $Sym (G)$ with $S_{n}$ . Theorem Cayley's theorem then realises $G$ as a subgroup of $S_{n}$ .∎

Remark. The embedding is rarely efficient. A group of order $n$ lands inside $S_{n}$ , which has $n!$ elements, so for $n = 12$ we are placing a group of order $12$ inside one of order $479, 001, 600$ . Better embeddings exist — the minimal degree of a faithful permutation representation is a genuine invariant, and for $Z /6 Z$ it is $5$ , not $6$ , since $Z /6 Z ≅ Z /2 Z \times Z /3 Z$ acts on $2 + 3$ points. What Cayley's theorem provides is existence, uniformly and with no hypotheses.

Example 3.44 (The Cayley embedding of $Z /4 Z$ ). Write out $Λ (1)$ for $G = Z /4 Z = {0, 1, 2, 3}$ , labelling the elements $0, 1, 2, 3$ as the points $1, 2, 3, 4$ respectively, and give the order of $Λ (1)$ as a permutation.

Solution.

$λ_{1}$ is "add $1$ ": it sends $0 \mapsto 1$ , $1 \mapsto 2$ , $2 \mapsto 3$ , $3 \mapsto 0$ .
In the labelling $0, 1, 2, 3 \leftrightarrow 1, 2, 3, 4$ this is the $4$ -cycle $(1234) \in S_{4}$ .
A $4$ -cycle has order $4$ , matching $∣1∣ = 4$ in $Z /4 Z$ , as Proposition The order of an image divides the order of the element requires for an injective map.
The image $Λ (G) = {e, (1234), (13) (24), (1432)}$ is a cyclic subgroup of order $4$ inside a group of order $24$ .

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Proposition 3.45 (The regular representation is fixed-point free). For $g \neq = e$ , the permutation $λ_{g}$ has no fixed point, and all of its cycles have the same length, namely the order of $g$ .

Proof. If $λ_{g} (x) = x$ then $gx = x$ , and cancelling $x$ on the right gives $g = e$ ; so for $g \neq = e$ there are no fixed points. The cycle of $λ_{g}$ containing $x$ is ${x, gx, g^{2} x, \dots}$ , and $g^{k} x = x$ holds if and only if $g^{k} = e$ , i.e. if and only if $∣ g ∣$ divides $k$ . So every cycle has length exactly $∣ g ∣$ .∎

Corollary 3.46 (Groups of even order contain an odd permutation). Let $∣ G ∣ = n$ be even and let $g \in G$ have order $2$ . Then $λ_{g}$ is a product of $n /2$ transpositions, so $sgn (λ_{g}) = (- 1)^{n /2}$ . In particular, if $n \equiv 2 (mod 4)$ then $G$ has a subgroup of index $2$ .

Proof. By Proposition The regular representation is fixed-point free, $λ_{g}$ has all cycles of length $2$ , and they partition the $n$ points, so there are $n /2$ of them and $sgn (λ_{g}) = (- 1)^{n /2}$ .

If $n \equiv 2 (mod 4)$ then $n /2$ is odd and $sgn (λ_{g}) = - 1$ . (Such a $g$ exists by Cauchy's theorem, or by pairing each element with its inverse.) Then $sgn \circ Λ : G \to {\pm 1}$ is a homomorphism that is surjective, so its kernel has index $2$ by Corollary The counting formula.∎

Exercises for §3.5 →

3.6Automorphisms

An automorphism is an isomorphism from a group to itself: a relabelling of the elements that leaves the multiplication table unchanged. Collecting them gives a new group attached to $G$ , and its size measures how much internal symmetry $G$ has.

Theorem 3.47 ( $Aut (G)$ is a group). Let $Aut (G)$ be the set of automorphisms of $G$ . Then $Aut (G)$ is a group under composition, and it is a subgroup of $Sym (G)$ .

Proof. Every automorphism is in particular a bijection $G \to G$ , so $Aut (G) \subseteq Sym (G)$ , and composition in $Sym (G)$ is associative. It remains to verify the subgroup conditions.

Non-empty. $id_{G}$ is a bijective homomorphism, so $id_{G} \in Aut (G)$ .

Closed under composition. If $α, β \in Aut (G)$ then $α \circ β$ is a homomorphism by Proposition Composites and inverses of homomorphisms and a bijection as a composite of bijections.

Closed under inverses. If $α \in Aut (G)$ then $α^{- 1}$ is a bijective homomorphism, again by Proposition Composites and inverses of homomorphisms.

By the subgroup criterion, $Aut (G) \leq Sym (G)$ , and in particular it is a group.∎

Definition 3.48 (Inner automorphism). For $g \in G$ , conjugation by $g$ is the map

γ_{g} : G \to G, γ_{g} (x) = gx g^{- 1} .

An automorphism of this form is called inner, and $Inn (G) = {γ_{g} : g \in G}$ . Automorphisms not of this form, or more precisely the elements of the quotient $Aut (G) / Inn (G)$ , are called outer.

Theorem 3.49 (Inner automorphisms, the centre, and the outer quotient). Let $G$ be a group. The map $γ : G \to Aut (G)$ , $g \mapsto γ_{g}$ , is a homomorphism with image $Inn (G)$ and kernel $Z (G)$ , the centre of $G$ . Consequently

$Inn (G) \leq Aut (G)$ and $Inn (G) ≅ G / Z (G)$ ;
$Inn (G) ⊴ Aut (G)$ ;
the quotient $Out (G) := Aut (G) / Inn (G)$ is a group.

Proof. Each $γ_{g}$ is an automorphism. It is a homomorphism: $γ_{g} (x y) = gx y g^{- 1} = (gx g^{- 1}) (g y g^{- 1}) = γ_{g} (x) γ_{g} (y)$ , the middle step inserting $g^{- 1} g$ . It is bijective with inverse $γ_{g^{- 1}}$ , since $γ_{g^{- 1}} (γ_{g} (x)) = g^{- 1} gx g^{- 1} g = x$ .

$γ$ is a homomorphism. For all $x$ ,

γ_{g} (γ_{h} (x)) = g (h x h^{- 1}) g^{- 1} = (g h) x (g h)^{- 1} = γ_{g h} (x),

so $γ_{g} \circ γ_{h} = γ_{g h}$ , i.e. $γ (g h) = γ (g) γ (h)$ .

Image and kernel. The image is $Inn (G)$ by definition, and it is a subgroup of $Aut (G)$ by Proposition The image is a subgroup; preimages are subgroups. The kernel is

ker γ = {g : gx g^{- 1} = x for all x} = {g : gx = xg for all x} = Z (G) .

Theorem First isomorphism theorem now gives $G / Z (G) ≅ Inn (G)$ , which is (1).

(2) Normality. Let $α \in Aut (G)$ and $g \in G$ . For every $x \in G$ ,

(α \circ γ_{g} \circ α^{- 1}) (x) = α (g α^{- 1} (x) g^{- 1}) = α (g) x α (g)^{- 1} = γ_{α (g)} (x),

using that $α$ is a homomorphism and $α (α^{- 1} (x)) = x$ . Hence $α γ_{g} α^{- 1} = γ_{α (g)} \in Inn (G)$ , so $Inn (G)$ is stable under conjugation by every automorphism and is therefore normal in $Aut (G)$ .

(3) A quotient by a normal subgroup is a group, so $Out (G) = Aut (G) / Inn (G)$ is defined.∎

The identity $α γ_{g} α^{- 1} = γ_{α (g)}$ is worth memorising: it says that relabelling the group by $α$ turns "conjugate by $g$ " into "conjugate by $α (g)$ ", which is exactly what one would want a relabelling to do. It is also the reason $Inn (G)$ is normal while a randomly chosen subgroup of $Aut (G)$ is not.

Corollary 3.50 (Abelian groups have no inner automorphisms). $G$ is abelian if and only if $Inn (G)$ is trivial; in that case $Out (G) = Aut (G)$ . Moreover, if $G / Z (G)$ is cyclic then $G$ is abelian.

Proof. $Inn (G)$ is trivial exactly when $Z (G) = G$ by Theorem Inner automorphisms, the centre, and the outer quotient, which is exactly the statement that $G$ is abelian.

For the last claim, suppose $G / Z (G) = ⟨ g Z (G)⟩$ is cyclic. Every $x \in G$ then lies in some coset $g^{k} Z (G)$ , so $x = g^{k} z$ with $z \in Z (G)$ . Given $x = g^{k} z$ and $y = g^{ℓ} w$ with $z, w$ central,

x y = g^{k} z g^{ℓ} w = g^{k + ℓ} z w = g^{ℓ} w g^{k} z = y x,

so $G$ is abelian.∎

Pitfall. The corollary has a famous consequence: $G / Z (G)$ is never cyclic and non-trivial. In particular a non-abelian group cannot have $[G : Z (G)] = p$ for a prime $p$ . Students often try to build a non-abelian group whose centre has index $3$ ; no such group exists.

Now the computational heart of the section. For cyclic groups the automorphism group is completely determined, and the answer is a familiar arithmetic object.

Theorem 3.51 ( $Aut (Z / n Z) ≅ (Z / n Z)^{\times}$ ). For every $n \geq 1$ ,

Aut (Z / n Z) ≅ (Z / n Z)^{\times},

the group of units modulo $n$ under multiplication. The isomorphism sends the unit $u$ to the map $σ_{u} (x) = ux mod n$ .

Proof. Write $C = Z / n Z$ additively, generated by $1$ .

Every automorphism has the form $σ_{u}$ . Let $α \in Aut (C)$ and put $u = α (1)$ . By Theorem Homomorphisms out of a cyclic group, $α (x) = α (x \cdot 1) = xu$ for every $x$ , so $α = σ_{u}$ .

$σ_{u}$ is an automorphism if and only if $u$ is a unit. For any $u$ , $σ_{u} (x + y) = u (x + y) = ux + u y$ , so $σ_{u}$ is always an endomorphism. Suppose $u$ is a unit, with $uv \equiv 1 (mod n)$ . Then $σ_{v}$ is a two-sided inverse for $σ_{u}$ , since $σ_{v} (σ_{u} (x)) = vux = x$ ; so $σ_{u}$ is bijective. Conversely, suppose $σ_{u}$ is bijective. Then it is surjective, so there is $v$ with $uv \equiv 1 (mod n)$ , i.e. $u$ is a unit. (Equivalently: $σ_{u}$ is injective iff $ker σ_{u} = {x : ux \equiv 0}$ is trivial, and that set is $⟨ n / g cd (u, n)⟩$ , which is trivial iff $g cd (u, n) = 1$ .)

The correspondence is a homomorphism. Define $Ψ : (Z / n Z)^{\times} \to Aut (C)$ by $Ψ (u) = σ_{u}$ . Then for all $x$ ,

(σ_{u} \circ σ_{v}) (x) = u (vx) = (uv) x = σ_{uv} (x),

so $Ψ (uv) = Ψ (u) Ψ (v)$ : $Ψ$ is a homomorphism.

Bijective. Surjectivity is the first paragraph together with the second. For injectivity, if $σ_{u} = id$ then evaluating at $x = 1$ gives $u \equiv 1 (mod n)$ , so $ker Ψ$ is trivial and Theorem Injectivity is detected by the kernel applies.

Hence $Ψ$ is an isomorphism.∎

Corollary 3.52 (The order of the automorphism group of a cyclic group). $Aut (Z / n Z) = φ (n)$ , where $φ$ is Euler's totient function. In particular, for a prime $p$ , $Aut (Z / p Z) ≅ Z / (p - 1) Z$ has order $p - 1$ .

Proof. By Theorem $Aut (Z / n Z) ≅ (Z / n Z)^{\times}$ , the order is $∣ (Z / n Z)^{\times} ∣$ , which is the number of residues coprime to $n$ , namely $φ (n)$ , as established in Modular Arithmetic. For $n = p$ prime every non-zero residue is a unit, so the order is $p - 1$ ; that this group is cyclic is the existence of a primitive root modulo $p$ .∎

Example 3.53 ( $Aut (Z /10 Z)$ ). Find the order of $Aut (Z /10 Z)$ , list its elements, and identify the group.

Solution.

By Corollary The order of the automorphism group of a cyclic group, the order is $φ (10) = φ (2) φ (5) = 1 \cdot 4 = 4$ .
The units modulo $10$ are ${1, 3, 7, 9}$ , giving $σ_{1} = id$ , $σ_{3}, σ_{7}, σ_{9}$ .
To identify the group, compute an order: $3^{2} = 9$ , $3^{3} = 27 \equiv 7$ , $3^{4} \equiv 21 \equiv 1 (mod 10)$ . So $σ_{3}$ has order $4$ and the group is cyclic: $Aut (Z /10 Z) ≅ Z /4 Z$ .
Note $∣ Aut (G) ∣ = 4 < 10 = ∣ G ∣$ : an automorphism group is usually much smaller than the group, and is never isomorphic to it for cyclic groups of order $> 2$ other than by accident.

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Example 3.54 ( $Aut (Z /8 Z)$ is not cyclic). Determine $Aut (Z /8 Z)$ up to isomorphism.

Solution.

$∣ Aut (Z /8 Z) ∣ = φ (8) = 4$ , with units ${1, 3, 5, 7}$ .
Square each unit modulo $8$ : $3^{2} = 9 \equiv 1$ , $5^{2} = 25 \equiv 1$ , $7^{2} = 49 \equiv 1$ . Every non-identity element has order $2$ .
A group of order $4$ with no element of order $4$ is the Klein four-group, by Example Distinguishing two groups of order 4. So $Aut (Z /8 Z) ≅ Z /2 Z \times Z /2 Z$ .
Contrast with Example $Aut (Z /10 Z)$ : both automorphism groups have order $4$ , but they are not isomorphic. The totient gives the size, never the structure.

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Example 3.55 ( $Aut (Z)$ ). Determine $Aut (Z)$ , where $Z$ is the additive group of integers.

Solution.

By Theorem Homomorphisms out of a cyclic group, an endomorphism of $Z$ is $x \mapsto ux$ for a unique $u = ϕ (1) \in Z$ .
Such a map is surjective if and only if $1$ is a multiple of $u$ , i.e. $u \in {1, - 1}$ ; both of these are also injective.
So $Aut (Z) = {id, x \mapsto - x}$ has order $2$ , and $Aut (Z) ≅ Z /2 Z$ .
This is consistent with the finite case: the "units of $Z$ " are $\pm 1$ , exactly the invertible multipliers.

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Example 3.56 ( $Aut (S_{3})$ and $Out (S_{3})$ ). Compute $∣ Inn (S_{3}) ∣$ and use it to determine $∣ Aut (S_{3}) ∣$ , given that $∣ Aut (S_{3}) ∣ \leq 6$ .

Solution.

$Z (S_{3}) = {e}$ : a non-identity permutation of three points commutes with everything only if it is the identity, as $(12)$ and $(13)$ do not commute and neither centralises the other.
By Theorem Inner automorphisms, the centre, and the outer quotient, $Inn (S_{3}) ≅ S_{3} / Z (S_{3}) = S_{3} / {e} ≅ S_{3}$ , so $∣ Inn (S_{3}) ∣ = 6$ .
Since $Inn (S_{3}) \leq Aut (S_{3})$ and $∣ Aut (S_{3}) ∣ \leq 6$ , we get $∣ Aut (S_{3}) ∣ = 6$ and $Aut (S_{3}) = Inn (S_{3}) ≅ S_{3}$ .
Hence $Out (S_{3}) = Aut (S_{3}) / Inn (S_{3})$ is trivial: every automorphism of $S_{3}$ is conjugation by an element of $S_{3}$ .

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Definition 3.57 (Characteristic subgroup). A subgroup $H \leq G$ is characteristic in $G$ , written $H char G$ , if $α (H) = H$ for every $α \in Aut (G)$ .

Proposition 3.58 (Characteristic implies normal, and characteristic is transitive). Let $H \leq G$ .

If $H char G$ then $H ⊴ G$ .
If $K char H$ and $H ⊴ G$ , then $K ⊴ G$ .

Proof. (1) Every inner automorphism $γ_{g}$ is an automorphism, so $γ_{g} (H) = g H g^{- 1} = H$ for all $g \in G$ , which is normality.

(2) Let $g \in G$ . Since $H ⊴ G$ , the map $γ_{g}$ restricts to an automorphism of $H$ (it maps $H$ into $H$ , and $γ_{g^{- 1}}$ provides the inverse). As $K$ is characteristic in $H$ , this restricted automorphism carries $K$ to $K$ , i.e. $g K g^{- 1} = K$ . Hence $K ⊴ G$ .∎

Part (2) is the reason the notion exists. Normality is famously not transitive — a normal subgroup of a normal subgroup need not be normal in the whole group — and characteristic is the strengthening that repairs the failure. The centre $Z (G)$ and the commutator subgroup $[G, G]$ are characteristic in any $G$ , since any automorphism must carry commuting elements to commuting elements and commutators to commutators.

Intuition. An automorphism is a way of repainting the elements of $G$ so that the multiplication table, read in the new colours, is identical to the old one. Inner automorphisms are the repaintings the group can perform on itself, by conjugating; $Out (G)$ measures the symmetries that are not visible from inside. That $Inn (G) ≅ G / Z (G)$ says the amount of self-repainting a group can do is exactly the amount by which it fails to be abelian: an abelian group is completely rigid from the inside and all of its symmetry is outer.

Exercises for §3.6 →

Common Mistakes

Checking the wrong condition. The one requirement is $ϕ (ab) = ϕ (a) ϕ (b)$ , with the left product taken in $G$ and the right in $H$ . Verifying $ϕ (e) = e$ or $ϕ (a^{- 1}) = ϕ (a)^{- 1}$ proves nothing; those are consequences, by Proposition A homomorphism preserves the identity, inverses and powers.
Calling a bijection an isomorphism. A bijection between the underlying sets carries no algebra. $Z /4 Z$ and the Klein four-group are in bijection and are not isomorphic.
Forgetting well-definedness. Any map defined on cosets by a formula in a representative — including the $\overline{ϕ}$ of Theorem First isomorphism theorem — must be shown independent of the representative. This is the crux of the proof, not a formality, and it is the step that fails if you quotient by a proper subgroup of the kernel.
Assuming a homomorphism is injective or surjective. Most are neither, and the interesting ones usually collapse something. The kernel measures the first failure and the image the second.
Quotienting by a non-normal subgroup. $G / H$ is a group only when $H ⊴ G$ . By Theorem Every normal subgroup is a kernel this is not a restriction that can be worked around — it is exactly the condition for $H$ to be a kernel.
Writing $HK / K ≅ H / K$ . The second isomorphism theorem gives $HK / K ≅ H / (H \cap K)$ ; the intersection is essential, and $K$ need not be contained in $H$ at all.
Applying the third isomorphism theorem without nesting. $(G / N) / (M / N) ≅ G / M$ requires $N \subseteq M$ . Without it, $M / N$ is not even a subset of $G / N$ .
Confusing $Aut (G)$ with $G$ . They are different groups and usually of different orders: $∣ Aut (Z /8 Z) ∣ = 4$ , and $Aut (Z / p Z)$ has order $p - 1$ , not $p$ .
Reading the totient as the structure. $∣ Aut (Z / n Z) ∣ = φ (n)$ gives the size only. $φ (8) = φ (10) = 4$ , yet one of these automorphism groups is Klein and the other is cyclic.
Expecting $Inn (G)$ to be all of $Aut (G)$ . It is for $S_{3}$ , and it is not in general; for an abelian group $Inn (G)$ is trivial while $Aut (G)$ can be large.

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