Homomorphisms and Isomorphism Theorems

A homomorphism $\phi : G \to H$ is a function satisfying $\phi(ab) = \phi(a)\phi(b)$ for all $a, b \in G$.

Proposition 5.1. If $\phi : G \to H$ is a homomorphism, then:

1. $\phi(e_G) = e_H$.
2. $\phi(a^{-1}) = \phi(a)^{-1}$ for all $a \in G$.
3. $\phi(a^n) = \phi(a)^n$ for all $n \in \mathbb{Z}$.

Proof. (1) $\phi(e_G) = \phi(e_G \cdot e_G) = \phi(e_G)\phi(e_G)$So by cancellation in $H$ $\phi(e_G) = e_H$. (2) $e_H = \phi(e_G) = \phi(aa^{-1}) = \phi(a)\phi(a^{-1})$So $\phi(a^{-1}) = \phi(a)^{-1}$. $\blacksquare$

The kernel of $\phi$ is $\ker(\phi) = \{g \in G : \phi(g) = e_H\}$.

The image of $\phi$ is $\mathrm{im}(\phi) = \{\phi(g) : g \in G\}$.

Theorem 5.2. $\ker(\phi) \trianglelefteq G$ and $\mathrm{im}(\phi) \leq H$.

Proof. We show $\ker(\phi)$ is a normal subgroup.

• $\phi(e) = e_H$So $e \in \ker(\phi)$.
• If $a, b \in \ker(\phi)$Then $\phi(ab^{-1}) = \phi(a)\phi(b)^{-1} = e_H e_H = e_H$So $ab^{-1} \in \ker(\phi)$.
• If $a \in \ker(\phi)$ and $g \in G$Then $\phi(gag^{-1}) = \phi(g)\phi(a)\phi(g)^{-1} = \phi(g)e_H\phi(g)^{-1} = e_H$ so $gag^{-1} \in \ker(\phi)$.

Thus $\ker(\phi) \trianglelefteq G$. The …/1-number-and-algebra/3_proof-and-logic that $\mathrm{im}(\phi) \leq H$ is straightforward. $\blacksquare$

Theorem 5.3 (First Isomorphism Theorem). If $\phi : G \to H$ is a surjective homomorphism, then

$G / \ker(\phi) \cong H$

More generally (even if $\phi$ is not surjective), $G / \ker(\phi) \cong \mathrm{im}(\phi)$.

Proof. Define $\overline{\phi} : G/\ker(\phi) \to H$ by $\overline{\phi}(g \ker(\phi)) = \phi(g)$. This is well-defined: if $g \ker(\phi) = g" \ker(\phi)$Then $g'^{-1}g \in \ker(\phi)$ So $\phi(g'^{-1}g) = e_H$Giving $\phi(g) = \phi(g')$. It is a homomorphism: $\overline{\phi}((a\ker(\phi))(b\ker(\phi))) = \overline{\phi}(ab\ker(\phi)) = \phi(ab) = \phi(a)\phi(b) = \overline{\phi}(a\ker(\phi))\overline{\phi}(b\ker(\phi))$. It is injective: $\overline{\phi}(g\ker(\phi)) = e_H \Rightarrow \phi(g) = e_H \Rightarrow g \in \ker(\phi) \Rightarrow g\ker(\phi) = \ker(\phi)$. By construction, $\mathrm{im}(\overline{\phi}) = \mathrm{im}(\phi)$. $\blacksquare$

Theorem 5.4 (Second Isomorphism Theorem). If $N \trianglelefteq G$ and $H \leq G$Then $HN \leq G$ $N \trianglelefteq HN$, $H \cap N \trianglelefteq H$And

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

Proof. Define $\phi : H \to HN/N$ by $\phi(h) = hN$. This is a homomorphism (since $N$ is normal). It is surjective: any element of $HN/N$ has the form $hnN = hN = \phi(h)$. Its kernel is $\{h \in H : hN = N\} = \{h \in H : h \in N\} = H \cap N$. By the first isomorphism theorem, $H/(H \cap N) \cong HN/N$. $\blacksquare$

Theorem 5.5 (Third Isomorphism Theorem). If $K \trianglelefteq N \trianglelefteq G$ with $K \trianglelefteq G$Then

$(G/K)/(N/K) \cong G/N$

Proof. Define $\phi : G/K \to G/N$ by $\phi(gK) = gN$. Well-defined: $gK = g'K$ implies $g^{-1}g' \in K \subseteq N$So $gN = g'N$. Surjective and $\ker(\phi) = N/K$. Apply the first Isomorphism theorem. $\blacksquare$

Problem. Show that $\mathbb{R}^* / \mathbb{R}^+ \cong \mathbb{Z}/2\mathbb{Z}$.

Solution. Define $\phi : \mathbb{R}^* \to \{1, -1\} \cong \mathbb{Z}/2\mathbb{Z}$ by $\phi(x) = \mathrm{sgn}(x)$. This is a homomorphism since $\mathrm{sgn}(xy) = \mathrm{sgn}(x)\mathrm{sgn}(y)$. It is surjective. Its kernel is $\{x \in \mathbb{R}^* : \mathrm{sgn}(x) = 1\} = \mathbb{R}^+$. By the first isomorphism theorem, $\mathbb{R}^* / \mathbb{R}^+ \cong \mathbb{Z}/2\mathbb{Z}$. $\blacksquare$

Example 5.6 (Trivial Homomorphism). For any groups $G, H$The map $\phi : G \to H$ defined by $\phi(g) = e_H$ for all $g \in G$ is a homomorphism (the trivial homomorphism). Its kernel is $G$ And its image is $\{e_H\}$.

Example 5.7 (Sign Homomorphism). The sign map $\mathrm{sgn} : S_n \to \{1, -1\}$ is a surjective Homomorphism with kernel $A_n$. By the first isomorphism theorem, $S_n / A_n \cong \mathbb{Z}/2\mathbb{Z}$.

Example 5.8 (Determinant Homomorphism). The determinant $\det : GL_n(\mathbb{R}) \to \mathbb{R}^*$ Is a surjective group homomorphism with kernel $SL_n(\mathbb{R})$. By the first isomorphism theorem, $GL_n(\mathbb{R}) / SL_n(\mathbb{R}) \cong \mathbb{R}^*$.

Example 5.9 (Conjugation Homomorphism). For a fixed $g \in G$The map $c_g : G \to G$ defined by $c_g(x) = gxg^{-1}$ is an automorphism of $G$ (an inner automorphism). It is a homomorphism since $c_g(xy) = gxyg^{-1} = (gxg^{-1})(gyg^{-1}) = c_g(x)c_g(y)$. It is bijective with inverse $c_{g^{-1}}$.

Problem. Define $\phi : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ by $\phi(k) = k + n\mathbb{Z}$. Verify that $\phi$ is a surjective homomorphism and identify the quotient.

Theorem 5.6 (Correspondence Theorem / Fourth Isomorphism Theorem). Let $\phi : G \to H$ be a Surjective homomorphism with $K = \ker(\phi)$. Then there is an inclusion-preserving bijection

$\{\mathrm{subgroups\ of\ } G \mathrm{\ containing\ } K\} \longleftrightarrow \{\mathrm{subgroups\ of\ } H\}$

Given by $U \mapsto \phi(U)$ with inverse $V \mapsto \phi^{-1}(V)$. This bijection satisfies:

1. $U \trianglelefteq G$ if and only if $\phi(U) \trianglelefteq H$.
2. $[G : U] = [H : \phi(U)]$ for all $U \supseteq K$.
3. $G/U \cong H/\phi(U)$ when $U \trianglelefteq G$.

Proof. Define $\Phi(U) = \phi(U)$ and $\Psi(V) = \phi^{-1}(V) = \{g \in G : \phi(g) \in V\}$.

• $\Phi \circ \Psi = \mathrm{id}$: $\Phi(\Psi(V)) = \phi(\phi^{-1}(V)) = V$ (since $\phi$ is surjective).
• $\Psi \circ \Phi = \mathrm{id}$ on subgroups containing $K$: if $U \supseteq K$Then $\Psi(\Phi(U)) = \phi^{-1}(\phi(U)) = U$ (since $\ker(\phi) = K \subseteq U$).

For normality: if $U \trianglelefteq G$Then for any $h \in H$ and $u \in U$ Write $h = \phi(g)$. Then $h\phi(u)h^{-1} = \phi(g)\phi(u)\phi(g)^{-1} = \phi(gug^{-1}) \in \phi(U)$ Since $gug^{-1} \in U$. Conversely, if $\phi(U) \trianglelefteq H$Then for any $g \in G$ and $u \in U$ $\phi(gug^{-1}) = \phi(g)\phi(u)\phi(g)^{-1} \in \phi(U)$So $gug^{-1} \in \phi^{-1}(\phi(U)) = U$.

For the index: $\phi$ restricts to a surjection $U \to \phi(U)$ with kernel $K$ So $|U/K| = |\phi(U)|$Giving $|G|/|U| = |H|/|\phi(U)|$. $\blacksquare$

An automorphism of $G$ is an isomorphism $\phi : G \to G$. The set of all automorphisms of $G$ Forms a group under composition, denoted $\mathrm{Aut}(G)$.

For each $g \in G$The inner automorphism $c_g : G \to G$ is defined by $c_g(x) = gxg^{-1}$. The set of inner automorphisms $\mathrm{Inn}(G) = \{c_g : g \in G\}$ is a normal subgroup of $\mathrm{Aut}(G)$.

Proposition 5.7. $\mathrm{Inn}(G) \cong G/Z(G)$.

Proof. Define $\psi : G \to \mathrm{Aut}(G)$ by $\psi(g) = c_g$. This is a homomorphism: $\psi(gh) = c_{gh}$And $c_{gh}(x) = ghx(gh)^{-1} = g(hxh^{-1})g^{-1} = c_g(c_h(x)) = (c_g \circ c_h)(x)$. The image is $\mathrm{Inn}(G)$. The kernel is $\{g \in G : c_g = \mathrm{id}\} = \{g \in G : gxg^{-1} = x \ \mathrm{for\ all\ x} \in G\} = Z(G)$. By the first isomorphism theorem, $\mathrm{Inn}(G) \cong G/Z(G)$. $\blacksquare$

Example. $\mathrm{Aut}(S_3) \cong S_3$. Since $Z(S_3) = \{e\}$We have $\mathrm{Inn}(S_3) \cong S_3$. Since $|\mathrm{Aut}(S_3)| \leq |S_3|! = 6$ (automorphisms permute the three elements of order $2$), And $\mathrm{Inn}(S_3)$ already has $6$ elements, we get $\mathrm{Aut}(S_3) = \mathrm{Inn}(S_3) \cong S_3$.

Example. $\mathrm{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^*$The group of units Modulo $n$. An automorphism of $\mathbb{Z}/n\mathbb{Z}$ is determined by where it sends $1$And $1$ can map to any generator, i.e., any $[k]$ with $\gcd(k, n) = 1$.

Example. $\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z}) \cong (\mathbb{Z}/8\mathbb{Z})^* = \{1, 3, 5, 7\} \cong V_4$. The four automorphisms are $x \mapsto x$, $x \mapsto 3x$, $x \mapsto 5x$, $x \mapsto 7x$. Note that $3^2 = 9 \equiv 1 \pmod{8}$So every non-identity automorphism has order $2$.

Definition. Let $H$ and $K$ be groups and let $\phi : K \to \mathrm{Aut}(H)$ be a homomorphism. The semidirect product $H \rtimes_\phi K$ is the set $H \times K$ with the group operation

$(h_1, k_1)(h_2, k_2) = (h_1 \cdot \phi(k_1)(h_2), k_1 k_2)$

When $\phi$ is the trivial homomorphism, this reduces to the direct product $H \times K$.

Proposition 5.8. In $G = H \rtimes_\phi K$The subgroup $H' = \{(h, e_K)\}$ is normal in $G$ And $K' = \{(e_H, k)\}$ is a subgroup with $H' \cap K' = \{(e_H, e_K)\}$ and $H'K' = G$.

Proof. Conjugation: $(e_H, k)(h, e_K)(e_H, k)^{-1} = (e_H \cdot \phi(k)(h), k)(e_H, k^{-1}) = (\phi(k)(h), e_K) \in H'$. So $H' \trianglelefteq G$. The remaining claims are immediate. $\blacksquare$

Example. $D_n = \mathbb{Z}/n\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$ where the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{Z}/n\mathbb{Z}$ sends $k \mapsto -k$ (inversion). Explicitly: $(a, 0)(b, 0) = (a + b, 0)$, $(a, 0)(b, 1) = (a + b, 1)$ $(a, 1)(b, 0) = (a - b, 1)$, $(a, 1)(b, 1) = (a - b, 0)$.

Problem. Show that there are exactly two groups of order $6$: $\mathbb{Z}/6\mathbb{Z}$ and $S_3$.

Problem. Classify all groups of order $21 = 3 \cdot 7$.