MHB O(G)=56. Sylow 2 subgroup has all its elements of order 2.

[caffeinemachine]

Let $G$ be a group of order $56$ having at least $7$ elements of order $7$.
1) Prove that $G$ has only one Sylow $2$-subgroup $P$.
2) All elements of $P$ have order $2$.

The first part is easy since it follows that the number of Sylow $7$-subgroups is $8$.
I got stuck on part 2. From part 1 we conclude that $P\triangleleft G$. So if $Q$ is any Sylow $7$ subgroup then $G=PQ=QP$. But I am getting nowhere with this. Please help.

 

[Deveno]

since P is normal we can let any sylow 7-subgroup Q act on it by conjugation.

consider the kernel of this action: it must be a subgroup of Q, which is cyclic of prime order. if the kernel is all of Q, then every element induces the identity map:

qpq^-1 = p, for all p in P, and all q in Q.

but this means that PQ = G is abelian (since any two generators for P and Q commute), contradicting the non-normality of Q.

this means that a generator x in Q induces a 7-cycle in P, that is: all non-identity elements of P are conjugate. since conjugates all have the same order, it must be that every element of P has order 2 (since it has at least one element of order 2 by Cauchy's theorem).