Does the symmetric group $S_{n+m}$ have a subgroup isomorphic to $S_n\times S_m$?

Question

Does the symmetric group $S_{n+m}$ have a subgroup isomorphic to $S_n\times S_m$?

What I've been trying is just manually trying to find a subgroup that's isomorphic to $S_n\times S_m$, but I was wondering if I could use one of the isomorphism theorems.

Answer 1

The group $S_{n+m}$ can be realized as the group of bijections from the set $\{1,\ldots,m+n\}$ to itself. Now consider the subgroup of bijections that map the subsets $\{1,\ldots,m\}$ and $\{m+1,\ldots,m+n\}$ to themselves.

Answer 2

Hint: if $H$ and $K$ are subgroups of a group $G$, $H \cap K = \{e\}$, and every element of $H$ commutes with every elements of $K$, then $HK = \{ hk : h\in H, k \in K\}$ is a subgroup of $G$ which is isomorphic to the direct product $H \times K$.

Answer 3

Consider the sets

$$S=\{\sigma\in S_{m+n}\mid \sigma(i)=i,\,\forall i\in\{m+1,m+2,\dots, m+n\}\}$$

and

$$T=\{\tau\in S_{m+n}\mid \tau(j)=j,\,\forall j\in\{1,2,\dots, m\}\}.$$

Clearly $S, T\le S_{m+n}$. Moreover, the elements of $S$ commute with the elements of $T$ and vice versa. Also $S\cong S_m$ and $T\cong S_n$. The (internal) direct product of two subgroups of a group is a subgroup of that group. It is routine to check, then, that $S\times T\cong S_m\times S_n$.