Prove that $S_4$ has no subgroup isomorphic to $Q_8$.

We know that $H$ contains the $4$-cycle $\sigma=(a\,b\,c\,d)$, because $H$ contains all $4$-cycles. Squaring $\sigma$ gives the double transposition $(a\,c)(b\,d)$, which lies in $H$ by closure under products. This is a generic double transposition (i.e. a "product of two $2$-cycles"), so $H$ contains all the double transpositions, i.e. $H$ contains the Klein $4$-group $V_4$.

Now $H$ contains at least $6$ elements (from the $4$-cycles), as well as $4$ new elements: those from $V_4$. Thus $\vert H \vert \geq 10 >8$, and this is the claim that the author wants.