Question: Prove that $S_4$ has no subgroup isomorphic to $Q_8$?
My attempt: If $Q_8$ isomorphic to subgroup of $S_4$ then there exists a homomorphism from $f: S_4\rightarrow Q_8$ such that,
$\frac{S_4}{\ker f}≈Q_8$ (Am I correct?)
this implies $|\ker f|$ must be equal to $3$. But $S_4$ has no normal subgroup of order $3$ and so... statement follows...
Is my attempt correct?
The quaternion group has six elements of order four, and they all have the same square. $S_4$ also has six elements of order four, the six four-cycles, but they don't all have the same square. So, $S_4$ can't have a subgroup isomorphic to the quaternion group.