For any finite group $G$ of order $n$ show that there exists two subgroups $H_1, H_2 \leq S_n$ isomorphic to $G$ with $h_1h_2 = h_2h_1$

Question

For any finite group $G$ of order $n$ show that there exists two subgroups $H_1, H_2 \leq S_n$ isomorphic to $G$ with $h_1h_2 = h_2h_1$ for all $h_1 \in H_1, h_2 \in H_2$.

Here is my attempt so far: Let $\sigma, \tau \in S_n$ be defined via cycle notation by $\sigma = (1, 2, ..., n)$ and $\tau = (n, n-1, ..., 1)$. The order of both $\sigma$ and $\tau$ is $n$, and furthermore $\tau = \sigma^{-1}$. We can then define

$$ H_1 = \langle \sigma \rangle, \quad H_2 = \langle \tau \rangle $$

so that $H_1$ and $H_2$ are both cyclic subgroups of $S_n$ of order $n$. Then we have

$$ H_1 \cong \mathbb{Z}/n\mathbb{Z} \cong H_2, $$

and if $h_1 \in H_1, h_2 \in H_2$ then $h_1h_2 = \sigma^i\tau^j = \sigma^i(\sigma^{-1})^j = \sigma^i\sigma^{-j} \sigma^{i-j} = \sigma^{-j}\sigma^i = \tau^j\sigma^i = h_2h_1$. This seems to be a solution to the problem, provided that the original group $G$ being considered is cyclic (and therefore isomorphic to $\mathbb{Z}/n\mathbb{Z}$). How do I proceed without the assumption that $G$ is a cyclic group of order $n$ however?

Answer 1

Hint: consider the permutations induced by left and right multiplication.

Answer 2

(I'm assuming you have at hands that, if $|G|=n$, then $\operatorname{Sym}(G)\cong S_n$.)

Take $H_1:=\{l_a, a\in G\}$ and $H_2:=\{r_a, a\in G\}$, where for every $a\in G$: \begin{alignat}{1} &\operatorname{Sym}(G)\ni l_a\colon g\mapsto ag \\ &\operatorname{Sym}(G)\ni r_a\colon g\mapsto ga^{-1} \\ \end{alignat} Prove that $l\colon a\mapsto l_a$ and $r\colon a\mapsto r_a$ are injective homomorphisms of $G$ into $\operatorname{Sym}(G)$, whence $H_1,H_2\cong G$. Moreover, given any $a,b\in G$, for every $g\in G$ we get: \begin{alignat}{1} (l_ar_b)(g) &= l_a(r_b(g)) \\ &= l_a(gb^{-1}) \\ &= agb^{-1} \\ &= (ag)b^{-1} \\ &= r_b(ag) \\ &= r_b(l_a(g)) \\ &= (r_bl_a)(g) \\ \end{alignat} and hence $l_ar_b=r_bl_a$ for every $a,b\in G$ (which is yours "$h_1h_2=h_2h_1$ for all etc."). This holds for infinite $G$ either, provided that you replace $S_n$ with $\operatorname{Sym}(G)$. Moreover, it may worth noting that $H_1\cap H_2\cong Z(G)$ and (as a corollary of your claim) $H_1H_2\le\operatorname{Sym}(G)$.

Just to getting familiar with these isomorphic images of $G$ into $\operatorname{Sym}(G)$ (we'll come up indeed to a trivial result), in the finite case these two latter facts tell us that: $$|H_1H_2|=\frac{n^2}{|Z(G)|}\mid n! \tag1$$ So, if $G$ is centerless, then from $(1)$ follows $n\mid (n-1)!$, namely $n$ is composite, whence the result that $G$ of prime order can't be centerless (which, as said, is by far trivial being in that case $G$ cyclic, and hence abelian).

A less trivial result with $H_1$ and $H_2$ is the following: an element of $H_1H_2$ (which is a bijection of $G$) is operation-preserving (namely an automorphism of $G$) if and only if, for every $x,y\in G$: \begin{alignat}{1} &(l_ar_b)(xy)=((l_ar_b)(x))((l_ar_b)(y))&&\iff \\ &axyb^{-1}=axb^{-1}ayb^{-1}&&\iff \\ &e=b^{-1}a&&\iff \\ &b=a&&\iff\\ &l_ar_b=l_ar_a\in\operatorname{Inn}(G) \end{alignat} whence: $$\operatorname{Inn}(G)=\operatorname{Aut}(G)\cap H_1H_2$$ thus making somehow plausible calling "inner" the automorphisms by conjugation (after all, isn't $H_1H_2$ "the greater area of $G$" in $\operatorname{Sym}(G)$?)