Exercise 3.5.11 ($S_4$ has no subgroup isomorphic to $Q_8$)

Proof.

Assume for the sake of contradiction that there is a subgroup $H$ of $S_4$ isomorphic to $Q_8$. By Exercise 1.3.4, $S_4$ has exactly $6$ elements of order $4$ (the six $4$-cycles). But $Q_8$ has $6$ elements of order $4$ by Exercise 1.5.1 ( $\pm i,\pm j,\pm k$). Therefore $H$ contains all the $4$-cycles. Then $H$ contains all the squares of these $4$-cycles, which are $(13)(24)$, $(12)(34)$ and $(14)(23)$. So $H$ contains $3$ elements of order $2$ (and $6$ elements of order $4$). Therefore $|H|\ge 9$, in contradiction with $|H|=8$.

So $S_4$ has no subgroup isomorphic to $Q_8$. □