$S_6$ can't have a subgroup isomorphic to $S_3\times S_4$ How to show that $S_6$ can't have a subgroup isomorphic to $S_3\times S_4$?
Please help me. I'm clueless.
 A: A quick check shows that $S_6$ has no elements of order$~12$ (considering a cycle decoposition, the orders that occur are obtained as least common multiple of the parts of a partition of$~6$, which gives $\{1,2,3,4,5,6\}$ as set of possible orders). But $S_3\times S_4$ does have an element of order$~12$.
A: $|S_3\times S_4|=3!.4!=6!/5=|S_6|/5$. 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?).
A: Show that an element of order three in $ S_6 $ cannot commute with one of order four.
A: Altenratively, you can just look at the orders of elements. What is the maximal order of an element in $S_{6}$? What is the maximal order of an element in $S_{3}\times S_{4}$? 
SPOILER

 The order of an element of $S_{n}$ is the least common multiple of the orders of its disjoint cycles. In $S_{6}$ the largest possible order is $6$ - coming from a 6-cycle or a product of a 2-cycle and a 3-cycle. In a similar way, the largest possible orders of elements of $S_{3}$ and $S_{4}$ are 3 and 4 respectively (arising from a 3-cycle and a 4-cycle). Thus the largest possible order of an element of $S_{3}\times S_{4}$ is $3\times 4=12$. Since by the work above, no element of $S_{6}$ has order 12, we can deduce that no subgroup of $S_{6}$ is isomorphic to $S_{3}\times S_{4}$.

