Chapter 06

Hard

Group Homomorphisms & Isomorphism Theorems

00 · Symbol Glossary

$\varphi: G \to H$Homomorphism from G to H

A function between two groups that respects the group operation: $\varphi(ab)=\varphi(a)\varphi(b)$ . The single defining requirement of this chapter — everything else follows from it.

$\ker\varphi$Kernel of a homomorphism

$\ker\varphi = \{g\in G : \varphi(g) = e_H\}$ — everything that collapses to the identity. Measures exactly how far $\varphi$ is from being injective.

$\mathrm{im}\,\varphi$Image of a homomorphism

$\mathrm{im}\,\varphi = \{\varphi(g) : g\in G\} \subseteq H$ — everything actually hit by $\varphi$ . Measures exactly how far $\varphi$ is from being surjective (onto $H$ ).

$\cong$Isomorphic

$G \cong H$ means there is a bijective homomorphism between them — the two groups are structurally identical, differing only in the names of their elements. The strongest possible relationship between two groups.

$e_G,\ e_H$Identity of G, identity of H

Subscripts distinguish which group's identity is meant, since a homomorphism deals with two groups at once and their identities are, a priori, different elements of different sets.

$HN$Product of two subsets

$HN = \{hn : h\in H, n\in N\}$ . Not generally a subgroup unless at least one of $H,N$ is normal — a fact made precise in Section 07.

01 · Homomorphisms

Chapter 05 built one structure-preserving relationship between groups: the quotient $G/N$ . Homomorphisms generalize this idea to any structure-preserving map between any two groups — quotients will turn out to be a special case.

Definition — Group Homomorphism

Let $G,H$ be groups. A function $\varphi: G\to H$ is a homomorphism if:

\varphi(ab) = \varphi(a)\varphi(b) \qquad \text{for all } a,b\in G

Note: the multiplication on the left is $G$ 's operation; on the right, $H$ 's.

Example — The sign homomorphism

Define $\text{sgn}: S_n \to \{1,-1\}$ (with $\{1,-1\}$ a group under ordinary multiplication) by $\text{sgn}(\sigma)=1$ if $\sigma$ is even, $-1$ if odd. Since parity adds when permutations compose (even∘even=even, even∘odd=odd, odd∘odd=even — exactly matching $1\times1=1$ , $1\times(-1)=-1$ , $(-1)\times(-1)=1$ ), $\text{sgn}(\sigma\tau)=\text{sgn}(\sigma)\text{sgn}(\tau)$ — a genuine homomorphism.

Example — Reduction mod n

Define $\pi:\mathbb{Z}\to\mathbb{Z}_n$ by $\pi(a) = [a]$ (the congruence class of $a$ ). Then $\pi(a+b) = [a+b] = [a]+[b] = \pi(a)+\pi(b)$ — exactly Theorem 1.5's well-definedness statement, now recognized as the homomorphism property.

Common mistake — Confusing a homomorphism with an arbitrary function

The function $f:\mathbb{Z}\to\mathbb{Z}$ , $f(n)=n+1$ , is not a homomorphism of $(\mathbb{Z},+)$ : $f(a+b) = a+b+1$ , but $f(a)+f(b) = (a+1)+(b+1) = a+b+2$ — different, unless coincidentally $1=2$ . A homomorphism is a very specific kind of function, not just "any" map between two groups' underlying sets.

02 · Basic Consequences

Theorem 6.1 — Homomorphisms Preserve Identity and Inverses

For any homomorphism $\varphi: G\to H$ : (a) $\varphi(e_G) = e_H$ ; (b) $\varphi(a^{-1}) = \varphi(a)^{-1}$ for every $a\in G$ .

Proof

(a) $\varphi(e_G) = \varphi(e_G e_G) = \varphi(e_G)\varphi(e_G)$ . Multiply both sides on the left by $\varphi(e_G)^{-1}$ (which exists in $H$ ): $e_H = \varphi(e_G)$ .

(b) $\varphi(a)\varphi(a^{-1}) = \varphi(aa^{-1}) = \varphi(e_G) = e_H$ by part (a). So $\varphi(a^{-1})$ satisfies the defining property of $\varphi(a)^{-1}$ ; by uniqueness of inverses (Theorem 1.2), $\varphi(a^{-1})=\varphi(a)^{-1}$ .

Theorem 6.2 — Homomorphic Images of Subgroups Are Subgroups

If $\varphi: G\to H$ is a homomorphism and $K\le G$ , then $\varphi(K) = \{\varphi(k):k\in K\} \le H$ .

Proof

Nonempty: $e_H = \varphi(e_G) \in \varphi(K)$ by Theorem 6.1(a), since $e_G\in K$ .

Closure: for $\varphi(k_1),\varphi(k_2)\in\varphi(K)$ , $\varphi(k_1)\varphi(k_2) = \varphi(k_1k_2)$ , and $k_1k_2\in K$ (closure in $K$ ), so this is in $\varphi(K)$ .

Inverses: $\varphi(k)^{-1} = \varphi(k^{-1})$ by Theorem 6.1(b), and $k^{-1}\in K$ , so this is in $\varphi(K)$ .

By Theorem 3.1, $\varphi(K)\le H$ .

03 · Kernel and Image

Definition — Kernel and Image

For $\varphi: G\to H$ : $\ker\varphi = \{g\in G : \varphi(g)=e_H\}$ , and $\mathrm{im}\,\varphi = \varphi(G) = \{\varphi(g):g\in G\}$ .

Theorem 6.3 — The Kernel Is a Normal Subgroup of G

$\ker\varphi \trianglelefteq G$ .

Proof

Subgroup: nonempty, since $\varphi(e_G)=e_H$ (Theorem 6.1a) puts $e_G\in\ker\varphi$ . Closure: if $\varphi(a)=e_H$ and $\varphi(b)=e_H$ , then $\varphi(ab)=\varphi(a)\varphi(b) = e_He_H=e_H$ , so $ab\in\ker\varphi$ . Inverses: if $\varphi(a)=e_H$ , then $\varphi(a^{-1}) = \varphi(a)^{-1} = e_H^{-1}=e_H$ (Theorem 6.1b), so $a^{-1}\in\ker\varphi$ . By Theorem 3.1, $\ker\varphi\le G$ .

Normal: take any $g\in G$ and $k\in\ker\varphi$ . Then:

\varphi(gkg^{-1}) = \varphi(g)\varphi(k)\varphi(g^{-1}) = \varphi(g)\,e_H\,\varphi(g)^{-1} = \varphi(g)\varphi(g)^{-1} = e_H

using $\varphi(k)=e_H$ and $\varphi(g^{-1})=\varphi(g)^{-1}$ (Theorem 6.1b). So $gkg^{-1}\in\ker\varphi$ for every $g,k$ , meaning $g(\ker\varphi)g^{-1}\subseteq\ker\varphi$ for every $g$ ; applying this with $g^{-1}$ gives the reverse inclusion. Hence $\ker\varphi\trianglelefteq G$ .

By Theorem 6.2, $\mathrm{im}\,\varphi = \varphi(G) \le H$ is also automatic (applying the theorem to $K=G$ ).

Example — Kernel of the sign homomorphism

$\ker(\text{sgn}) = \{\sigma \in S_n : \text{sgn}(\sigma)=1\} = A_n$ — the even permutations. This matches Chapter 05's finding that $A_n \trianglelefteq S_n$ , now explained: $A_n$ is normal precisely because it's the kernel of a homomorphism, not merely because it happens to have index 2.

04 · Injectivity and the Kernel

Theorem 6.4 — Injective iff Trivial Kernel

A homomorphism $\varphi: G\to H$ is injective if and only if $\ker\varphi = \{e_G\}$ .

Proof

( $\Rightarrow$ ) Suppose $\varphi$ injective. If $g\in\ker\varphi$ , then $\varphi(g)=e_H=\varphi(e_G)$ ; injectivity forces $g=e_G$ . So $\ker\varphi\subseteq\{e_G\}$ , and since $e_G\in\ker\varphi$ always, $\ker\varphi=\{e_G\}$ .

( $\Leftarrow$ ) Suppose $\ker\varphi=\{e_G\}$ . If $\varphi(a)=\varphi(b)$ , then $\varphi(a)\varphi(b)^{-1}=e_H$ , i.e. $\varphi(a)\varphi(b^{-1})=e_H$ (Theorem 6.1b), i.e. $\varphi(ab^{-1})=e_H$ . So $ab^{-1}\in\ker\varphi=\{e_G\}$ , giving $ab^{-1}=e_G$ , i.e. $a=b$ . So $\varphi$ is injective.

A one-element check replaces an all-pairs check

Ordinarily, proving a function injective means checking " $\varphi(a)=\varphi(b)\implies a=b$ " for every pair $a,b$ . Theorem 6.4 reduces this, for homomorphisms specifically, to checking a single set — the kernel — for a single property: does it contain anything besides $e_G$ ? This shortcut is used constantly from here on.

05 · Isomorphisms

Definition — Isomorphism

A homomorphism $\varphi: G\to H$ that is also a bijection is called an isomorphism. If one exists, $G$ and $H$ are isomorphic, written $G\cong H$ .

Isomorphic groups are indistinguishable as groups

Every structural property phrased purely in terms of the group operation — order, abelian-ness, element orders, number of subgroups of each order, and so on — transfers exactly across an isomorphism. $G\cong H$ means "the same group wearing different labels," which is why $D_3 \cong S_3$ (both size 6, both nonabelian, matching element-order profiles) was asserted without further proof back in Chapter 02.

Example — Verifying $D_3 \cong S_3$

Define $\varphi: D_3 \to S_3$ by sending each symmetry to the permutation it induces on the triangle's vertices $\{1,2,3\}$ : $r\mapsto(1\,2\,3)$ , $s\mapsto(2\,3)$ (fixing vertex 1), and extend to all $6$ elements by composing. Composing two symmetries and then reading off vertex permutations gives the same answer as reading off vertex permutations first and then composing — so $\varphi$ is a homomorphism. Since both groups have order $6$ and $\varphi$ can be checked directly to hit all six permutations without repeats, $\varphi$ is bijective. $D_3\cong S_3$ .

06 · The First Isomorphism Theorem

This theorem is the payoff for building kernels, images, and quotients side by side: every homomorphism factors, essentially uniquely, into a quotient map followed by an isomorphism.

Theorem 6.5 — First Isomorphism Theorem

For any homomorphism $\varphi: G\to H$ :

G/\ker\varphi \;\cong\; \mathrm{im}\,\varphi

Proof

Write $K=\ker\varphi$ (normal by Theorem 6.3, so $G/K$ is a genuine group by Theorem 5.3). Define $\bar\varphi: G/K \to \mathrm{im}\,\varphi$ by $\bar\varphi(gK) = \varphi(g)$ .

Well-defined. If $gK=g'K$ , then by Theorem 4.1, $g^{-1}g'\in K=\ker\varphi$ , so $\varphi(g^{-1}g')=e_H$ , i.e. $\varphi(g)^{-1}\varphi(g')=e_H$ (Theorem 6.1b applied inside this computation), giving $\varphi(g')=\varphi(g)$ . So $\bar\varphi(gK)$ doesn't depend on which representative $g$ of the coset was used.

Homomorphism. $\bar\varphi(gK\cdot g'K) = \bar\varphi(gg'K) = \varphi(gg') = \varphi(g)\varphi(g') = \bar\varphi(gK)\bar\varphi(g'K)$ .

Injective. If $\bar\varphi(gK)=e_H$ , then $\varphi(g)=e_H$ , so $g\in K$ , meaning $gK=eK$ — the identity coset. By Theorem 6.4 (applied to $\bar\varphi$ itself), trivial kernel means injective.

Surjective onto $\mathrm{im}\,\varphi$ . Every element of $\mathrm{im}\,\varphi$ has the form $\varphi(g)$ for some $g\in G$ , and $\bar\varphi(gK)=\varphi(g)$ hits it directly.

$\bar\varphi$ is a bijective homomorphism $G/K\to\mathrm{im}\,\varphi$ : an isomorphism. Hence $G/\ker\varphi\cong\mathrm{im}\,\varphi$ .

Example — Recovering $S_n/A_n \cong \{1,-1\}$

$\ker(\text{sgn}) = A_n$ , and $\mathrm{im}(\text{sgn}) = \{1,-1\}$ (sgn hits both values, e.g. sgn(e)=1, sgn of a transposition $=-1$ ). By Theorem 6.5:

S_n/A_n \;\cong\; \{1,-1\}

This instantly confirms $|S_n/A_n|=2$ , matching Lagrange's Theorem applied to $[S_n:A_n]=2$ from Chapter 05 — but now with an explicit isomorphism identifying which group of order 2 it is, not merely its size.

Example — $\mathbb{Z}/n\mathbb{Z} \cong \mathrm{im}$ of reduction mod n

For $\pi:\mathbb{Z}\to\mathbb{Z}_n$ (reduction mod $n$ ), $\ker\pi = n\mathbb{Z}$ (exactly the multiples of $n$ , which map to $[0]$ ) and $\pi$ is surjective onto $\mathbb{Z}_n$ (every class $[a]$ is hit by $\pi(a)$ ). Theorem 6.5 gives $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n$ — confirming, via the general machinery, exactly what Chapter 05 observed directly by hand.

07 · The Second and Third Isomorphism Theorems

Two further theorems, both consequences of the First, describe how quotients interact with subgroups and with each other.

Theorem 6.6 — Second Isomorphism Theorem

Let $H\le G$ and $N\trianglelefteq G$ . Then $HN=\{hn:h\in H,n\in N\}$ is a subgroup of $G$ , $H\cap N \trianglelefteq H$ , and:

HN/N \;\cong\; H/(H\cap N)

Proof

$HN\le G$ : nonempty ( $e=ee\in HN$ ). Closure: $(h_1n_1)(h_2n_2) = h_1h_2(h_2^{-1}n_1h_2)n_2$ ; since $N\trianglelefteq G$ , $h_2^{-1}n_1h_2\in N$ , so $(h_2^{-1}n_1h_2)n_2\in N$ , giving $(h_1n_1)(h_2n_2) = (h_1h_2)\cdot(\text{something in } N) \in HN$ . Inverses: $(hn)^{-1}=n^{-1}h^{-1} = h^{-1}(hn^{-1}h^{-1})$ , and $hn^{-1}h^{-1}\in N$ by normality, so $(hn)^{-1}\in HN$ . By Theorem 3.1, $HN\le G$ .

Define $\psi: H \to HN/N$ by $\psi(h) = hN$ . This is a homomorphism: $\psi(h_1h_2) = h_1h_2N = (h_1N)(h_2N) = \psi(h_1)\psi(h_2)$ , valid since $N\trianglelefteq HN$ (as $N\trianglelefteq G$ , normality holds against every element of $G$ , in particular every element of $HN$ ), so $HN/N$ is a genuine group by Theorem 5.3.

$\psi$ is surjective onto $HN/N$ : every coset in $HN/N$ has the form $(hn)N = hN$ (since $nN=N$ ) for some $h\in H$ , which is $\psi(h)$ .

$\ker\psi = H\cap N$ : $\psi(h)=N$ (the identity of $HN/N$ ) $\iff hN=N \iff h\in N$ . Combined with $h\in H$ (domain of $\psi$ ), this is exactly $h\in H\cap N$ . By Theorem 6.3, $\ker\psi=H\cap N \trianglelefteq H$ .

By the First Isomorphism Theorem applied to $\psi$ : $H/\ker\psi \cong \mathrm{im}\,\psi$ , i.e. $H/(H\cap N) \cong HN/N$ .

Theorem 6.7 — Third Isomorphism Theorem

Let $N,K\trianglelefteq G$ with $N\le K$ . Then $K/N \trianglelefteq G/N$ , and:

(G/N)\big/(K/N) \;\cong\; G/K

Proof

Define $\Psi: G/N \to G/K$ by $\Psi(gN) = gK$ .

Well-defined: if $gN=g'N$ , then $g^{-1}g'\in N \subseteq K$ , so $g^{-1}g'\in K$ , giving $gK=g'K$ .

Homomorphism: $\Psi(gN\cdot g'N) = \Psi(gg'N) = gg'K = (gK)(g'K) = \Psi(gN)\Psi(g'N)$ .

Surjective onto $G/K$ : every coset $gK$ is $\Psi(gN)$ .

$\ker\Psi = K/N$ : $\Psi(gN) = K$ (identity of $G/K$ ) $\iff gK=K \iff g\in K$ . The set of such cosets $gN$ (for $g\in K$ ) is exactly $K/N=\{kN : k\in K\}$ . By Theorem 6.3, $K/N=\ker\Psi \trianglelefteq G/N$ .

By the First Isomorphism Theorem: $(G/N)/\ker\Psi \cong \mathrm{im}\,\Psi$ , i.e. $(G/N)/(K/N) \cong G/K$ .

A recurring template

Every proof in this section follows the exact same three-step template: construct a natural map, identify its kernel, invoke Theorem 6.5. This is deliberate — the First Isomorphism Theorem is doing essentially all of the work every time; the Second and Third are really just clever choices of which map to feed into it.

08 · Exercises

EXERCISE 6.1

Apply Theorem 6.1 directly, using that $\det(I)=1$ is $GL_n(\mathbb{R})$ 's identity's image.

$\det: GL_n(\mathbb{R}) \to \mathbb{R}\setminus\{0\}$ satisfies $\det(AB)=\det(A)\det(B)$ — a homomorphism. By Theorem 6.1(a), $\det(I) = 1$ (the identity of $\mathbb{R}\setminus\{0\}$ under multiplication), confirming what's already known about determinants directly, but now as a forced consequence of the homomorphism property alone.

Confirm that $\det: GL_n(\mathbb{R}) \to \mathbb{R}\setminus\{0\}$ is a homomorphism, and use Theorem 6.1(a) to state what $\det(I)$ must equal.

EXERCISE 6.2

$\ker(\det)$ is exactly $SL_n(\mathbb{R})$ from Exercise 3.1. Apply Theorem 6.3 directly.

$\ker(\det) = \{A\in GL_n(\mathbb{R}) : \det A=1\} = SL_n(\mathbb{R})$ , exactly the subgroup from Exercise 3.1. By Theorem 6.3, the kernel of any homomorphism is normal: $SL_n(\mathbb{R}) \trianglelefteq GL_n(\mathbb{R})$ — a fact that would require direct conjugation computation with matrices to verify by hand, but follows immediately here.

Using $\det: GL_n(\mathbb{R})\to\mathbb{R}\setminus\{0\}$ and Theorem 6.3, show that $SL_n(\mathbb{R})=\{A:\det A=1\}$ is normal in $GL_n(\mathbb{R})$ .

EXERCISE 6.3

Apply the First Isomorphism Theorem to $\det: GL_n(\mathbb{R})\to\mathbb{R}\setminus\{0\}$ : what is its image?

$\det$ is surjective onto $\mathbb{R}\setminus\{0\}$ (for any nonzero real $c$ , the diagonal matrix with $c$ in one entry and $1$ s elsewhere has determinant $c$ ). By Theorem 6.5:

$GL_n(\mathbb{R})/SL_n(\mathbb{R}) \;\cong\; \mathbb{R}\setminus\{0\}$

State the conclusion of the First Isomorphism Theorem applied to $\det: GL_n(\mathbb{R}) \to \mathbb{R}\setminus\{0\}$ .

EXERCISE 6.4

Use Theorem 6.4: compute the kernel directly, then decide if it's trivial.

$\varphi:\mathbb{Z}_6\to\mathbb{Z}_3$ by $\varphi([a]_6) = [a]_3$ (reduce further mod 3). Check well-defined and homomorphism properties hold (similar to Theorem 1.5's argument). $\ker\varphi = \{[a]_6 : a\equiv0\pmod3\} = \{0,3\}$ — not just $\{0\}$ .

By Theorem 6.4, since $\ker\varphi\neq\{e\}$ , $\varphi$ is not injective. (In fact $|\ker\varphi|=2$ , so exactly two elements of $\mathbb{Z}_6$ map to $0$ .)

Let $\varphi: \mathbb{Z}_6 \to \mathbb{Z}_3$ send $[a]_6 \mapsto [a]_3$ . Find $\ker\varphi$ and determine, using Theorem 6.4, whether $\varphi$ is injective.

EXERCISE 6.5

Apply the First Isomorphism Theorem to the $\varphi$ from Exercise 6.4, using $\ker\varphi=\{0,3\}$ and surjectivity onto $\mathbb{Z}_3$ .

$\varphi$ from Exercise 6.4 is surjective onto $\mathbb{Z}_3$ (every class mod 3 is hit). By Theorem 6.5:

$\mathbb{Z}_6/\{0,3\} \;\cong\; \mathbb{Z}_3$

Sanity check via order: $|\mathbb{Z}_6/\{0,3\}| = 6/2 = 3 = |\mathbb{Z}_3|$ . ✓ — consistent, and now with an explicit isomorphism rather than just matching cardinalities.

Using the map from Exercise 6.4, state what the First Isomorphism Theorem gives for $\mathbb{Z}_6/\ker\varphi$ , and check the orders match.

09 · Chapter Summary

Concept	Statement
Homomorphism	$\varphi:G\to H$ with $\varphi(ab)=\varphi(a)\varphi(b)$
Preserves identity, inverses	$\varphi(e_G)=e_H$ ; $\varphi(a^{-1})=\varphi(a)^{-1}$ (Thm 6.1)
Image of a subgroup	$\varphi(K)\le H$ when $K\le G$ (Thm 6.2)
Kernel	$\ker\varphi = \{g:\varphi(g)=e_H\}$ ; always normal in $G$ (Thm 6.3)
Injective test	$\varphi$ injective $\iff \ker\varphi=\{e_G\}$ (Thm 6.4)
Isomorphism	Bijective homomorphism; $G\cong H$ means structurally identical
First Isomorphism Theorem	$G/\ker\varphi \cong \mathrm{im}\,\varphi$ (Thm 6.5)
Second Isomorphism Theorem	$HN/N \cong H/(H\cap N)$ (Thm 6.6)
Third Isomorphism Theorem	$(G/N)/(K/N) \cong G/K$ (Thm 6.7)

Next: Chapter 07 — Direct Products & Direct Sums builds larger groups out of smaller ones, giving the tools to construct (and recognize) groups like $\mathbb{Z}_2\times\mathbb{Z}_2$ that Exercise 5.2 hinted might differ from $\mathbb{Z}_4$ despite sharing the same order.