TITLE: $S_{n+1}$ not isomorphic to subgroup of $S_n \times S_n$
QUESTION [8 upvotes]: I've been asked to prove that there is no injective homomorphism from $S_{n+1}$ to $S_n \times S_n$ for $n\ge4$.  
This seems to me to follow from the fact that $S_{n+1}$ cannot be recognized as a direct product of two of its subgroups, essentially because it has only one normal subgroup.  Are there any ways to do this via order considerations, as this was my first impulse upon seeing the problem.

REPLY [3 votes]: Since $n \geq 4$, the only normal subgroups of $S_{n+1}$ are $\{(1)\}$, $A_{n+1}$ and $S_{n+1}$. This follows from the simplicity of $A_{n+1}$. Put $M=S_n \times \{(1)\}$. Then $M \lhd G:=S_n \times S_n$, with $|M|=n!$. Look at $K=M \cap S_{n+1}$. This is a normal subgroup of $S_{n+1}$, whence $K$ equals one of the three options mentioned before: If $K=(1)$, then $MS_{n+1}/M \cong S_{n+1}$. It follows that $(n+1)! \mid |G/M|=n!$, which is absurd.
If $K=A_{n+1}$, then $\frac{(n+1)!}{2} \mid |M|=n!$, which cannot be the case.
If $K=S_{n+1}$, then $S_{n+1} \subset M$ and we would have $(n+1)! \mid n!$, again a contradiction.