1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.