Group Homomorphisms and the Isomorphism Theorems

By [L3], for $u,v\in H$ choose $x,y\in G$ with $u=f(x)$ and $v=f(y)$; then $uv=f(xy)$ by [L2].

Therefore $f^{-1}(uv)=xy=f^{-1}(u)f^{-1}(v)$, so the inverse preserves the group operation.

Hence $f^{-1}:H\to G$ is a group homomorphism.

The image contains $e_H=f(e_G)$ and, for $f(x),f(y)\in \mathrm{im}f$, contains $f(x)f(y{)}^{-1}=f(x{y}^{-1})$; thus [L3] gives $\mathrm{im}f\le H$.

The kernel is a subgroup by the same calculation, and for $k\in \ker f$ one has $f(gk{g}^{-1})=f(g)e_{H}f(g{)}^{-1}=e_H$, so $g(\ker f)g^{-1}\subseteq \ker f$; applying this to $g^{-1}$ gives equality.

The conjugation calculation in step 2.1 completes both assertions.

If $f$ is injective and $g\in \ker f$, then $f(g)=e_H=f(e_G)$, so $g=e_G$.

Conversely, if $\ker f=\{e_G\}$ and $f(x)=f(y)$, then $f(y^{-1}x)=e_H$, whence $y^{-1}x=e_G$ and $x=y$.

The two implications prove the equivalence.

If $N=\ker f$ for a homomorphism, then $N$ is normal by [L1].

If $N$ is normal, [L2] supplies the quotient homomorphism $\pi$ and gives $\ker \pi =N$.

Thus normal subgroups are exactly kernels.

If $f(g)=f(h)$, then [L2] gives $f(h^{-1}g)=e_H$, so $h^{-1}g\in \ker f$ and [L3] gives $g\ker f=h\ker f$.

If $g\ker f=h\ker f$, then [L3] gives $h^{-1}g\in \ker f$, so $f(h{)}^{-1}f(g)=e_H$ and $f(g)=f(h)$.

This proves the stated equivalence.

Define $\bar{f}(gN):=f(g)$; if $gN=hN$, then $h^{-1}g\in N\subseteq \ker f$, so [L4] proves that this value is independent of the representative.

For cosets, $\bar{f}((gN)(hN))=f(gh)=f(g)f(h)$, and $f(g)=\bar{f}(\pi (g))$.

The surjectivity used in step 2.1 forces any such factor map to have these values, hence proves uniqueness.

By [L2] and [L1], $\bar{f}:G\mathrm{/}\ker f\to \mathrm{im}f$, $\bar{f}(g\ker f)=f(g)$, is a well-defined homomorphism; [L4] also gives representative independence directly.

Its image is all of $\mathrm{im}f$, and $\bar{f}(g\ker f)=e_H$ implies $f(g)=e_H$, hence $g\ker f=\ker f$; therefore its kernel is trivial.

The trivial-kernel conclusion of step 2.1 makes $\bar{f}$ an isomorphism.

The identity lies in $HN$; for $h_1{n}_1,h_2{n}_2\in HN$, put $m=n_1{n}_2^{-1}\in N$, so $(h_1{n}_1)(h_2{n}_2{)}^{-1}=h_{1}m{h}_2^{-1}=h_1{h}_2^{-1}(h_{2}m{h}_2^{-1})\in HN$.

Thus [L1] gives $HN\le G$; moreover for $h\in H$ and $x\in H\cap N$, both $hx{h}^{-1}\in H$ and $hx{h}^{-1}\in N$, so it lies in $H\cap N$.

The conjugation closure in step 2.1 gives $H\cap N⊴H$.

Restrict the quotient map $HN\to HN\mathrm{/}N$ to $\varphi :H\to HN\mathrm{/}N$, $\varphi (h)=hN$; [L1] and [L4] make this a homomorphism.

Its kernel is $\{h\in H:hN=N\}=H\cap N$, while every $hnN=hN$ shows that its image is $HN\mathrm{/}N$.

The kernel and image calculation in step 2.1 gives $H\mathrm{/}(H\cap N)\cong HN\mathrm{/}N$.

Since $K\subseteq N$, the subset $N\mathrm{/}K=\{nK:n\in N\}$ is a subgroup of $G\mathrm{/}K$ by the quotient product rule.

For $gK\in G\mathrm{/}K$ and $nK\in N\mathrm{/}K$, $(gK)(nK)(gK{)}^{-1}=(gn{g}^{-1})K$ belongs to $N\mathrm{/}K$ because $N⊴G$.

Thus the conjugation calculation gives $N\mathrm{/}K⊴G\mathrm{/}K$.

Define $\varphi :G\mathrm{/}K\to G\mathrm{/}N$ by $\varphi (gK)=gN$; it is well defined because $K\subseteq N$, and [L4] shows it is a homomorphism.

The map is onto and $\varphi (gK)=N$ exactly when $g\in N$, so $\ker \varphi =N\mathrm{/}K$.

The kernel and image calculation yields $(G\mathrm{/}K)\mathrm{/}(N\mathrm{/}K)\cong G\mathrm{/}N$.

For $N\le H\le G$, $H\mathrm{/}N=\pi [H]$ is a subgroup, while ${\pi }^{-1}(K)$ is a subgroup containing $\ker \pi =N$.

Surjectivity gives $\pi [{\pi }^{-1}(K)]=K$, and $N\le H$ gives ${\pi }^{-1}(\pi [H])=H$; both assignments therefore preserve inclusion and are inverse.

The image and preimage calculation of step 2.1 also preserves normality.

The identity map is an automorphism, and the composite of two automorphisms is again a bijective homomorphism.

By [L2], the inverse of every automorphism is an automorphism, while associativity comes from composition of functions.

Hence the closure and inverse properties in step 2.1 give a group structure on $\mathrm{Aut}(G)$.

Associativity gives $c_g(xy)=gxy{g}^{-1}=(gx{g}^{-1})(gy{g}^{-1})$, so $c_g$ is a homomorphism.

The map $c_{g^{-1}}$ is inverse to $c_g$ by cancellation.

Thus $c_g$ is a bijective homomorphism and hence an automorphism.

Define $C(g)=c_g$; for every $x$, $C(gh)(x)=ghx(gh{)}^{-1}=g(hx{h}^{-1})g^{-1}=(C(g)\circ C(h))(x)$.

Now $C(g)={\mathrm{id}}_G$ exactly when $gx{g}^{-1}=x$ for every $x$, equivalently $g\in Z(G)$, and its image is $\mathrm{Inn}(G)$ by definition.

Thus $C$ is a homomorphism with kernel $Z(G)$ and image $\mathrm{Inn}(G)$.

Apply [L2] to the conjugation homomorphism of [L1].

Substituting its kernel and image yields $G\mathrm{/}Z(G)\cong \mathrm{Inn}(G)$.

This is the asserted isomorphism.

For $\phi \in \mathrm{Aut}(G)$ and $c_g\in \mathrm{Inn}(G)$, direct evaluation gives $(\phi c_g{\phi }^{-1})(x)=\phi (g)x\phi (g{)}^{-1}=c_{\phi (g)}(x)$.

Thus conjugation by every element of $\mathrm{Aut}(G)$ carries $\mathrm{Inn}(G)$ into itself; applying the same statement to ${\phi }^{-1}$ gives equality.

The conjugation closure in step 2.1 proves normality.