Question

The number of involutions in G is |G|/4, and every right coset of a Sylow 2-subgroup S of G not contained in NG(S) contains exactly one involution.

Answer 1

The number of Sylow 2-subgroups of G is |G : NG(S)| = |G|/12 .Each of them contains three involutions, and any two of them have trivial intersection, byEvery Sylow 2-subgroup of G is self-centralising. Moreover, every two distinct Sylow 2-subgroups of G have trivial intersection, and G contains no elements of even order greater than 2.

by lemma:It follows that the number of involutions in G is 3|G:NG(S)| = |G|/4.

Now let g ∈ G\NG(S), and suppose that the coset Sg contains two involutions, say ug and vg, for distinct elements u, v ∈ S ∼= C2 × C2.( isomorphic)

Then since (ug) ^2 = 1 = (vg) ^2

, we have ugu = g ^−1 and vgv = g ^−1 , so uv commutes with g, and hence uvg has order 6 in G, contrary to the conclusion of Lemma 6. It follows that every right coset Sg with g /∈ NG(S) contains at most one involution. On the other hand, in NG(S) itself there is just one right coset of S containing involutions, namely the trivial coset S, which contains three involutions. Hence the remaining (|G| − 12)/4 right cosets of S in G contain the other |G|/4 − 3 involutions, which (by the earlier observation) implies that they contain exactly one each

we see that, |N : S| = 3 and

|N| = 3|S| = 12. It follows that N contains an element g of order 3, conjugation by which

cyclically permutes the three non-trivial elements of S, and the rest follows easily.