Problem 5.4.11 (If $G = HK$ where $H$ and $K$ are characteristic subgroups of $G$ with $H \cap K = 1$ then $\mathrm{Aut}(G) \simeq \mathrm{Aut}(H) \times \mathrm{Aut}(K)$)

Proof.

Note first that since $H$ and $K$ are characteristic subgroups of $G$, then $H⊴G$ and $K⊴G$. By Theorem 9, the map $\phi :H\times K\to G$ defined by $\phi (h,k)=\mathit{h𝑘}$ is an isomorphism, and so $G\simeq H\times K$. In particular every element of $H$ commutes with every element of $K$.

Let $\sigma \in \mathrm{Aut}(G)$ be an automorphism of $G=\mathit{𝐻𝐾}$. Since $H$ and $K$ are characteristic subgroups of $G$, then $\sigma (x)\in H$ for all $x\in H$, and $\sigma (x)\in K$ for all $x\in K$, so the maps

are well defined, and are automorphisms, so ${\sigma }_H\in \mathrm{Aut}(H)$ and ${\sigma }_K\in \mathrm{Aut}(K)$

Consider the map

Then

• $\chi$ is a homomorphism: if $\sigma ,\tau \in \mathrm{Aut}(G)$ and $\lambda =\sigma \circ \tau$, then for all $x\in H$, $({\sigma }_H\circ {\tau }_H)(x)=\sigma (\tau (x))={\lambda }_H(x)$, so ${\sigma }_H\circ {\tau }_H={\lambda }_H$ and similarly ${\sigma }_K\circ {\tau }_K={\lambda }_K$, then

  $$\begin{array}{llll}\hfill \chi (\sigma \circ \tau )& =\chi (\lambda )& \hfill & \\ \hfill & =({\lambda }_H,{\lambda }_K)& \hfill & \\ \hfill & =({\sigma }_H\circ {\tau }_H,{\sigma }_K\circ {\tau }_K)& \hfill & \\ \hfill & =({\sigma }_H,{\sigma }_K)({\tau }_H,{\tau }_K)& \hfill & \\ \hfill & =\chi (\sigma )\chi (\tau ),& \hfill & \end{array}$$

• $\chi$ is injective: if $\sigma \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\chi )$, then ${\sigma }_H={\mathrm{id}}_H$ and ${\sigma }_K={\mathrm{id}}_K$. Therefore, for all $h\in H$, $\sigma (h)={\sigma }_H(h)=h$ and $\sigma (k)={\sigma }_K(k)=k$. If $x\in G$, there are elements $h\in H$ and $k\in K$ such that $x=\mathit{h𝑘}$. Then

  $$\sigma (x)=\sigma (\mathit{h𝑘})=\sigma (h)\sigma (k)=\mathit{h𝑘}=x,$$

  so $\sigma ={\mathrm{id}}_G$. This shows $\ker <!--\; FUNCTION\; APPLICATION\; -->(\chi )=\{{\mathrm{id}}_G\}$, so $\chi$ is injective.

• $\chi$ is surjective: if $({\sigma }_1,{\sigma }_2)\in \mathrm{Aut}(H)\times \mathrm{Aut}(K)$, let $\sigma :G\to G$ be defined for every $x=\mathit{h𝑘}\in G$ by

  $$\sigma (x)={\sigma }_1(h){\sigma }_2(k).$$

  Since every element of $G$ is of the form $x=\mathit{h𝑘}$ in a unique way, $\sigma$ is well defined. Moreover, if $x=\mathit{h𝑘}\in G$ and $x^{\prime }=h^{\prime }{k}^{\prime }\in G$, where $h,h^{\prime }\in H$ and $k,k^{\prime }\in K$, then, since every element of $H$ commutes with every element of $K$, $x{x}^{\prime }=\mathit{h𝑘}{h}^{\prime }{k}^{\prime }=h{h}^{\prime }k{k}^{\prime }$, thus

  $$\begin{array}{llllll}\hfill \sigma (x{x}^{\prime })& ={\sigma }_1(h{h}^{\prime }){\sigma }_2(k{k}^{\prime })& \hfill & & \hfill & \\ \hfill & ={\sigma }_1(h){\sigma }_1(h^{\prime }){\sigma }_2(k){\sigma }_2(k^{\prime })& \hfill & & \hfill & \\ \hfill & ={\sigma }_1(h){\sigma }_2(k){\sigma }_1(h^{\prime }){\sigma }_2(k^{\prime })& \hfill (\text{since }{\sigma }_1(h^{\prime })\in H\text{ and }{\sigma }_2(k)\in K\text{ commute})& & \hfill & & \hfill \\ \hfill & =\sigma (x)\sigma (x^{\prime }),& \hfill & & \hfill & \end{array}$$

  so $\sigma :G\to G$ is a homomorphism.

  If $x=\mathit{h𝑘}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\sigma )$, then ${\sigma }_1(h)=1$ and ${\sigma }_2(k)=1$. Since ${\sigma }_1$ and ${\sigma }_2$ are injective, $h=k=1$, so $x=1$. Therefore $\sigma$ is injective.

  If $y\in G$, there are elements $h_1\in H$ and $k_1\in K$ such that $y=h_1{k}_1$. Put $h={\sigma }_1^{-1}(h_1)\in H$, $k={\sigma }_2^{-1}(k_1)\in K$, and $x=\mathit{h𝑘}$. Then $\sigma (x)={\sigma }_1(h){\sigma }_2(k)=h_1{k}_1=y$, so $\sigma$ is surjective.

  In conclusion $\sigma \in \mathrm{Aut}(G)$. Moreover, for all $h\in H$,

  $${\sigma }_H(h)=\sigma (h)={\sigma }_1(h){\sigma }_2(1)={\sigma }_1(h),$$

  so ${\sigma }_H={\sigma }_1$, and similarly ${\sigma }_K={\sigma }_2$. Therefore

  $$\chi (\sigma )=({\sigma }_H,{\sigma }_K)=({\sigma }_1,{\sigma }_2).$$

  This shows that $\chi$ is surjective.

In conclusion, $\chi$ is an isomorphism, and

We suppose now that $G$ is a finite abelian group.

Let $S_1,S_2,\dots ,S_r$ be the Sylow subgroups of $G$. By Exercise 10,

and $G\simeq H\times K$, where $H=S_1{S}_2\cdots S_{r-1}$ and $K=S_r$. (We use the same notations as in Exercise 10.)

Note that $H\cap K=\{1\}$, since $|H|=p_1^{{\alpha }_1}{p}_2^{{\alpha }_1}\cdots p_{r-1}^{{\alpha }_{r-1}}$ and $|K|=p_r^{{\alpha }_r}$ are relatively prime.

Every Sylow subgroup $S$ of $G$ is characteristic in $G$: If $|S|=p^{\alpha }$, then for every automorphism $\sigma \in \mathrm{Aut}(G)$, $|\sigma (S)|=|S|=p^{\alpha }$, so $\sigma (S)$ is a Sylow $p$-subgroup of $G$. Since $G$ is abelian, $S⊴G$, thus there is only one Sylow $p$-subgroup, so $\sigma (S)=S$. Consequently, $H=S_1{S}_2\cdots S_{r-1}$ is also characteristic: $\sigma (H)=\sigma (S_1)\sigma (S_2)\cdots \sigma (S_{r-1})=S_1{S}_2\cdots S_{r-1}=H$.

By the first part of this proof,

By induction,

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