Prove that $S_4$ has no subgroup isomorphic to $Q_8$. The question is to prove that $S_4$ has no subgroup isomorphic to $Q_8$. Here is an answer. But what "then $H$ also contains all products of two 2-cycles" means in that answer?
Thanks.
 A: 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.
A: You may check the argument computationally. $Q_8$, the quaternion group of order $8$, is the group SmallGroup(8,4) in the GAP Small Groups Library. In the following lines, I apply GAP to probe the case:
gap> s4:=SymmetricGroup(4);;
     e:=AllSubgroups(s4);

This returns $30$ subgroups, but only three of them are of order $8$. Let's find them as follows:
gap>  Filtered(e,t->Order(t)=8);


Group([ (1,4)(2,3), (1,3)(2,4), (3,4) ]),
Group([ (1,2)(3,4), (1,3)(2,4), (1,4) ]),
Group([ (1,2)(3,4), (1,4)(2,3), (2,4) ]),

Now, a simple code StructureDescription(...) tells us that the above subgroups are all $D_8$, not $Q_8$.
A: Consider a square with vertices labelled by $1,2,3,4$. The group of symmetries of this square is dihedral group of order $8$, and its elements can be written as permutations on four letters.
Hence the dihedral group of order $8$ is subgroup of $S_4$ and $S_4\leq S_5$.
Since the order of Sylow-$2$ subgroup of $S_5$ is $8$, the Sylow-$2$ subgroup should be dihedral group of order $8$. Hence quaternion group of order $8$ can not be in $S_5$ (o.w. it would be Sylow-$2$ subgroup, not isomorphic to dihedral group).
