$T$ is a minimal generating set for $S_n$. $T=${$(i,i+1),i=1,...,n−1$}.
For any fixed $x$ with $1≤x<n$, the set $W =$ $T$\ {$x,x+1 $} does not generate $S_n$.
$W$ generates some subgroup of $S_n$. I want to show that this subgroup is isomorphic to $S_x×S_{n-x}$.
First I let this subgroup be called $G$. So I need to show $G≅S_x×S_{n-x}$.
$W$ doesn't generate $S_n$ when we take out {$x,x+1$}. There's a partition $T_1$, $T_2$, where $T_1=${$1,....,x$},$T_2=$ {$x+1,....,n$}.
Is $S_x$ the $T_1$ and $S_{n-x}$ the $T_2$? And $S_x×S_{n-x}$ is the direct product where the operation is the permutation composition.
Now I need to show $G≅S_x×S_{n-x}$, so do I need to give an isomorphism to prove they are isomorphic, or is there something else I can do to prove it, maybe something about their order?
$T$ generates $S_n$ for every $n$. Hence $\{(1,2),...,(x-1,x)\}$ generates $S_{T_1}$ and $\{(x+1,x+2),...,(n-1,n)\}$ generates $S_{T_2}$. Let $G$ be the subgroup generated by $W\setminus \{(x,x+1)\}$. Then $T_1, T_2$ are invariant under the action of $G$. Hence every $g\in G$ induces a permutation $g_1$ of $S_{T_1}$ and a permutation $g_2\in S_{T_2}$. This gives an isomorphism $g\mapsto (g_1, g_2)$ from $G$ to $S_{T_1} \times S_{T_2}$.