Question. Let $H$ and $K$ be two finite groups that $(|H|,|K|)=1$.
(i)- Prove that every subgroup of $H \times K$ has the form $A \times B$ where $A$ is a subgroup of $H$ and $B$ is a subgroup of $K$.
(ii)- Prove $Aut(H \times K) \cong Aut(H)\times Aut(K)$
My attempt for (i).
Lemma 1. Let $G$ be a finite group and let $N$ a normal subgroup of $G$. Let $|N|=n$ and $|G:N|=m$, and assume $(m,n)=1$. Let $H$ any subgroup of $G$. Since $|H|||G|=nm$ with $m$ and $n$ relatively prime, there are unique integers $k$, $l$ such that $|H|=kl$ that $k|n$ and $l|m$. Then we have: $|H \cap N|=k$ and $|H:H \cap N|=l$.
Proof. As $N$ is a normal subgroup so $HN$ is subgroup of $G$.
$$
\begin{equation}
\begin{cases}
|HN|=\frac{|H||N|}{|H \cap N|}\\[2ex]
|G|=mn
\end{cases}
\xrightarrow{|HN|||G|}
\frac{kln}{|H \cap N|}|mn
\end{equation}\tag{1}\label{eq1}
$$
We know that $|H \cap N|||H|$, so \eqref{eq1} can be changed:
$$\frac{kl}{|H \cap N|}|m$$
Since $n$ and $m$ are relatively prime, and $\frac{k}{|H \cap N|}|m$ we can conclude that: $|H \cap N|=k$.$\blacksquare$
For part (i) we can assume these two normal subsets: $$\begin{cases} H'=H\times \{e_K\}\\[2ex] K'=\{e_H\}\times K \end{cases}$$ We now $H'$ and $K'$ are normal subgroups in $H \times K$ and these two subgroups have the lemma 1 property: $$\begin{cases} (|K'|,|H \times K:K'|)=(m,n)=1\\[2ex] (|H'|,|H \times K:H'|)=(n,m)=1 \end{cases}$$ Assume $T$ as subgroup in $H \times K$, and $|T|=kl$ that $k|n$ and $l|m$. As a result of lemma 1, we can conclude: $$\begin{cases} |K' \cap T|=l\\[2ex] |H' \cap T|=k \end{cases}$$
$K' \cap T$ is the elements of the form $(e_H,k)\in T$, and $H' \cap T$ are elements of the form $(h.e_K)\in T$. Since $T$ is subgroup $(H' \cap T).(K' \cap T)\subseteq T$. We can check that $\pi_1(H')$ and $\pi_2(K')$ are subgroups, respectively, in $H$ and $K$. Additionally, $(H' \cap T).(K' \cap T)=\pi_1(H') \times \pi_2(K') \subseteq T$. With counting elements of $\pi_1(H') \times \pi_2(K')=kl$, we can conclude $\pi_1(H') \times \pi_2(K')=T$.$\blacksquare$
(ii) For this part, we can make each isomorphism $g: H \times K \rightarrow H \times K $ as an element of $Aut(H)\times Aut(K)$. Consider $g^{-1}(H')$. It is a subgroup in $H \times K$. With the using (i), this subgroup should be $A \times B$ that $A$ subgroup in $H$ and $B$ a subgroup in $K$. As $g$ is an isomorphism: $$n=|H'|=|A||B| \xrightarrow{(|B|,n)=1} |B|=1 \implies A=H$$ Thus we can make $g_H:H \rightarrow H $, and $g_K: K \rightarrow K$ isomorphism: $$ g:H \times \{e_K\} \rightarrow H \times \{e_K\}\\[2ex] \begin{equation} \begin{cases} \begin{aligned} g_H : H &\rightarrow H\\ h &\rightarrow \pi_1(g(h\times e_K)) \end{aligned} \end{cases} \end{equation}\tag{2}\label{eq2}$$ $$ g: \{e_H\} \times K \rightarrow \{e_H\} \times K\\[2ex] \begin{equation} \begin{cases} \begin{aligned} g_K : K &\rightarrow K\\ k &\rightarrow \pi_2(g(e_H \times k)) \end{aligned} \end{cases} \end{equation}\tag{3}\label{eq3}$$ $g_K$ and $g_H$ are isomorphism, respectively, for $H$ and $K$. Moreover, we can write $g$ as following: $$\begin{cases} g(h \times e_K)=(g_H(h),e_K)\\[2ex] g(e_H \times k)=(e_H, g_K(k)) \end{cases}\\ \implies g(h \times k)=g((h \times e_K)(e_H \times k))=g(h \times e_K).g(e_H \times k)=(g_H(h),e_K).(e_H, g_K(k))=(g_H(h),g_K(k))$$ $\blacksquare$
This question has a very cute result, which I liked:)
You can conclude that $\phi(mn)=\phi(m).\phi(n)$ when $(m,n)=1$.
Questions:
- Is the proof correct?
- If you find out any other proof, please give me some hints too.
Thanks.
