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?

  • $\begingroup$ I think it would be clearer if you looked for homomorphisms $K \to {\rm Aut}(H)$. I know that ${\rm Aut}(H) \cong Z_6$, but that is not particularly helpful here. If $H = \langle x \rangle$, then the automorphisms of $H$ have the form $x \mapsto x^i$ for $1 \le i \le 6$. $\endgroup$ – Derek Holt Feb 19 '18 at 18:34
  • $\begingroup$ @DerekHolt I thought about that (that's where $Z_6$ came from) but in any case I don't know how this can be helpful. $\endgroup$ – user527831 Feb 19 '18 at 18:44

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 $G\cong\langle h,k\mid h^7=k^4=1,\;kh=h^{-1}k\rangle$.

  • $\begingroup$ Thanks. I don't understand the normalizer argument. We know that the order of $G$ is the product of the order of $N_G(K)$ and the number of conjugate subgroups of $K$, i.e., 7. Shouldn't then $N_G(K)$ have order 4? Why is it 7? And why does it divide $[G:H]$? $\endgroup$ – user527831 Feb 19 '18 at 20:26
  • $\begingroup$ $G$ acts on $K$ by conjugation. The size of the orbit (I.e. the number of Sylow 2-subgroups) equals the index of the stabilizer. The stabilizer is $N_G(K)$, so $7=[G:N_G(K)]$. Now, if $H\leq N_G(K)$ then $4=[G:H]=[G:N_G(K)][N_G(K):H]$. Hence 4|7, a contradiction. $\endgroup$ – David Hill Feb 19 '18 at 22:06
  • $\begingroup$ Can I just say istead that $khk^{-1}=h$ implies that the elements of $K$ commute with those of $H$, and also $KH=G$, so $G$ must be abelian? Also, it seems I don't understand why the orbit from your previous comment consists of Sylow 2-subgroups. This would be true if $G$ acted on the set of Sylow 2-subgroups, and not on some particular subgroup. $\endgroup$ – user527831 Feb 19 '18 at 22:35
  • $\begingroup$ Yes, you are right. $G$ acts on the set of conjugates of $K$. $\endgroup$ – David Hill Feb 20 '18 at 1:40

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