TITLE: $S_6$ can't have a subgroup isomorphic to $S_3\times S_4$
QUESTION [9 upvotes]: How to show that $S_6$ can't have a subgroup isomorphic to $S_3\times S_4$?
Please help me. I'm clueless.

REPLY [3 votes]: $|S_3\times S_4|=3!.4!=6!/5=|S_6|/5$. We can show that $S_6$ has no subgroup of order $3!4!$ (i.e. of index $5$). If $H\leq S_6$ has index $5$, then $S_6$ acts on the five cosets of $H$ in $S_6$ by multiplication (left/right-as one is convenient). Then there is a non-trivial homomorphism from $S_6$ to $S_5$ such that image has order $\geq 5$, i.e. the kernel has index $\leq 5$; contradiction (why?).