Prove that there is at most one non-abelian group of order $28$ all of whose Sylow $2$-subgroups are cyclic

Question

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?

Answer 1

As Derek Holt points out, it is better to regard $\mathrm{Aut}(H)\cong(\mathbb{Z}/7\mathbb{Z})^\times$, in which case the two possible homomorphisms $\mathbb{Z}/4\mathbb{Z}\to (\mathbb{Z}/7\mathbb{Z})^\times$ are $f(1)=1$ and $f(1)=6$ (i.e. $khk^{-1}=h$ or $khk^{-1}=h^6=h^{-1}$, where $K=\langle k\rangle$).

Suppose that $f(1)=1$. Then, $khk^{-1}=h$ for all $h\in H$ and $k\in K$ and, in turn, $hkh^{-1}=k$ for all $h\in H$ and $k\in K$. Therefore, $H\leq N_G(K)$. But, $[G:N_G(K)]$ divides $4=[G:H]$ which is impossible since $[G:N_G(K)]=7$.

Finally, we must have $f(1)=6$, and $G\cong\langle h,k\mid h^7=k^4=1,\;kh=h^{-1}k\rangle$.

Answer 2

By the counting argument, the $7$-Sylow subgroup $H$ is unique, hence normal. Moreover, said $K$ a $2$-Sylow subgroup, it is $H\cap K=1$. Therefore, any such a group arises as a nontrivial semidirect product $K\ltimes H$. Since $\operatorname {Aut}(H)\cong$ $\mathbb Z/6\mathbb Z$, if we prove that there is at most one nontrivial homomorphism $\varphi\colon \mathbb Z/4\mathbb Z\to \mathbb Z/6\mathbb Z$, then we are done. There are two options, only: $\ker\varphi$ is trivial or $\ker\varphi=\{0,2\}$. But the former is ruled out by Lagrange's theorem, as $4\nmid 6$. Therefore, the only possible option is that $0$ and $2$ are mapped to $0$, whereas $1,3$ (both of order $4$) are necessarily mapped to $3$, which is the only element of order $2$ in $\mathbb Z/6\mathbb Z$$^\dagger$. So, finally, there is at most one such a nontrivial homomorphism, namely at most one nonabelian group of order $28=2^27$ all of whose $2$-Sylow subgroups are cyclic.

(To prove that there is precisely one such a group, one should prove that the map singled out above is actually operation-preserving. That is true, but it goes beyond the scope of the question.)

---

$^{\dagger}$The order of the image of an element under a homomorphism divides the order of the element. And $\mathbb Z/6\mathbb Z$ hasn't got any element of order $4$.