## If the direct factors are characteristic in a direct product, then the automorphism group of the product is the product of automorphism groups

Prove that if $H$ and $K$ are groups such that $H \times 1$ and $1 \times K$ are characteristic in $H \times K$, then $\mathsf{Aut}(H \times K) \cong \mathsf{Aut}(H) \times \mathsf{Aut}(K)$. Deduce that if $G$ is an abelian group of finite order then $\mathsf{Aut}(G)$ is isomorphic to the direct product of the automorphism groups of its Sylow subgroups.

We define a mapping $\Phi : \mathsf{Aut}(H) \times \mathsf{Aut}(K) \rightarrow \mathsf{Aut}(H \times K)$ as follows: $\Phi(\varphi, \psi) = \varphi \times \psi$. We claim that $\Phi$ is an isomorphism.

1. ($\Phi$ is a homomorphism) Suppose all mappings are in the appropriate sets. Then $\Phi((\varphi_1,\psi_1)(\varphi_2,\psi_2)) = \Phi(\varphi_1 \circ \varphi_2, \psi_1 \circ \psi_2)$ $= (\varphi_1 \circ \varphi_2) \times (\psi_1, \circ \psi_2)$ $= (\varphi_1 \times \psi_1) \circ (\varphi_2 \times \psi_2)$ $= \Phi(\varphi_1, \psi_2) \circ \Phi(\varphi_2, \psi_2)$.
2. ($\Phi$ is injective) Suppose $\Phi(\varphi_1,\psi_1) = \Phi(\varphi_2,\psi_2)$. Then for all $h \in H$ and $k \in K$, we have $(\varphi_1(h), \psi_1(k)) = (\varphi_1 \times \psi_1)(h,k)$ $= (\varphi_2 \times \psi_2)(h,k)$ $= (\varphi_2(h), \psi_2(k))$. Since pairs are equal precisely when their corresponding entries are equal, we have $\varphi_1(h) = \varphi_2(h)$ and $\psi_1(k) = \psi_2(k)$ for all $h \in H$ and $k \in K$; thus $\varphi_1 = \varphi_2$ and $\psi_1 = \psi_2$, and we have $(\varphi_1,\psi_1) = (\varphi_2,\psi_2)$.
3. ($\Phi$ is surjective) Let $\chi \in \mathsf{Aut}(H \times K)$. Let $h \in H$; since $H \times 1 \leq H \times K$ is characteristic, $\chi(h,1) \in H \times 1$. Define a mapping $\chi_H : H \rightarrow H$ by $\chi(h,1) = (\chi_H(h), 1)$. We claim that $\chi_H$ is an automorphism of $H$.
1. ($\chi_H$ is a homomorphism) Note that $(\chi_H(h_1h_2), 1) = \chi(h_1h_2,1)$ $= \chi(h_1,1) \chi(h_2,1)$ $= (\chi_H(h_1),1) (\chi_H(h_2), 1)$ $= (\chi_H(h_1) \chi_H(h_2), 1)$. Comparing entries, we have $\chi_H(h_1h_2) = \chi_H(h_1) \chi_H(h_2)$ for all $h_1,h_2 \in H$.
2. ($\chi_H$ is injective) Suppose $\chi_H(h_1) = \chi_H(h_2)$. Then $\chi(h_1,1) = (\chi_H(h_1),1)$ $= (\chi_H(h_2),1)$ $= \chi(h_2,1)$. Since $\chi$ is injective, $h_1 = h_2$.
3. ($\chi_H$ is surjective) Let $h_1 \in H$. Since $\chi[H \times 1] = H \times 1$, there exists $h_2 \in H$ such that $\chi(h_2,1) = (h_1,1)$. Thus $h_1 = \chi_H(h_2)$.

Likewise, we can define a mapping $\chi_K : K \rightarrow K$ which is also an isomorphism.

Finally, note that $\chi(h,k) = \chi((h,1)(1,k))$ $\chi(h,1)\chi(1,k)$ $= (\chi_H(h),1)(1,\chi_K(k))$ $= (\chi_H(h), \chi_K(k))$ $= (\chi_H \times \chi_K)(h,k)$ Thus $\chi = \chi_H \times \chi_K$, and we have $\chi = \Phi(\chi_H, \chi_K)$.

Thus $\mathsf{Aut}(H \times K) \cong \mathsf{Aut}(H) \times \mathsf{Aut}(K)$.

As a lemma, we can extend this result by induction to finite direct products in which each factor is characteristic.

Lemma: Let $G_i$ be groups such that $G_{\{i\}}$ (c.f. §5.1 #2) is characteristic in $G = \times_{i=1}^n G_i$ for each $i$. Then $\mathsf{Aut}(\times_{i=1}^n G_i) \cong \times_{i=1}^n \mathsf{Aut}(G_i)$. Proof: We proceed by induction on $n$. The conclusion is trivial for $n=1$, and we proved the $n=2$ case above. Suppose the conclusion holds for some $n \geq 2$. Now $\times_{i=1}^{n+1} G_i \cong (\times{i=1}^n G_i) \times G_{n+1}$. Suppose $\varphi$ is an automorphism of $G$; then $\varphi[\times_{i=1}^n G_i] = \varphi[\prod_{i=1}^n G_{\{i\}}]$ $= \prod_{i=1}^n \varphi[G_{\{i\}}]$ $= \prod_{i=1}^n G_{\{i\}}$ $= \times_{i=1}^n G_i$, so that $\times_{i=1}^n G_i$ is characteristic. By the induction hypothesis and part (a) of this problem, we have $\mathsf{Aut}(\times_{i=1}^{n+1} G_i) \cong \times_{i=1}^{n+1} \mathsf{Aut}(G_i)$. $\square$

Now let $A$ be a finite abelian group. Note that every Sylow subgroup $P$ is normal, hence the unique subgroup of order $|P|$. Thus each Sylow subgroup is characteristic in $A$. In §5.4 #10, we saw that a finite abelian group is (isomorphic to) the direct product of its Sylow subgroups; the final conclusion then follows from the lemma.