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- #1

- Mar 10, 2012

- 835

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.