Let $G$ be a non-abelian group of order $28$ all of whose Sylow $2$-subgroups are cyclic. Show that there is at most one such group (up to isomorphism).
By Sylow's theorems, $G$ has a unique (normal) Sylow $7$-subgroup $H$ and $7$ Sylow $2$-subgroups (call one of them $K\simeq Z_4$). $K$ acts on $H$ by conjugation, which gives a homomorphism $f: Z_4\rightarrow Z_6$. There are 2 such homomorphisms: $f(1)=0,3$. I can't see what can I get from this. Is it a wrong way?