If $H\subseteq S_{n}$ and $K\subseteq S_{m}$ how can I then show that I can think of $H\times K$ as it was a subgroup of $S_{m+n}$?
two subgroups of $S_{n}$ and $S_{m}$
-
0Or for a different way of getting the same end result: conjugate $K$ within $S_{n+m}$ with the product of the $2$-cycles $(1,n+1)$, $(2,n+2)$,$\ldots$, $(m,n+m)$. But let $H$ be. – 2012-04-05
1 Answers
In hopes of getting this off the Unanswered list, here’s a hint expanding on Jyrki’s first comment.
$K$ is a group of permutations of the set $\{1,\dots,m\}$, so each $k\in K$ is a bijection $k:\{1,\dots,m\}\to\{1,\dots,m\}\;.$ For each $k\in K$ let $\hat k:\{n+1,\dots,n+m\}\to\{n+1,\dots,n+m\}:n+i\mapsto n+k(i)\;,$ and let $\widehat K=\{\hat k:k\in K\}$.
Show that $\widehat K$ is a group of permutations of $\{n+1,\dots,n+m\}$ and is isomorphic to $K$.
For each $\langle h,k\rangle\in H\times K$ let $g_{\langle h,k\rangle}:S_{n+m}\to S_{n+m}:i\mapsto\begin{cases} h(i),&\text{if }1\le i\le n\\ \hat k(i),&\text{if }n+1\le i\le n+m\;, \end{cases}$
and let $G=\{g_{\langle h,k\rangle}:\langle h,k\rangle\in H\times K\}$.
Show that $G$ is a subgroup of $S_{n+m}$ and is isomorphic to $H\times K$.