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The following is an exercise in Dummit and Foote (4.5.27) I'm doing for revision: let $G$ be a group of order 315 ($= 3^2 \cdot 5 \cdot7$) which has a normal Sylow 3-subgroup. Prove that $Z(G)$ contains a Sylow 3-subgroup of $G$ and deduce that $G$ is abelian ($Z(G)$ denotes the center of $G$, i.e. $Z(G) = \{g \in G: ga = ag \;\;\forall a \in G\}$

I'm not sure how to proceed. It looks like a possible approach would be to show that $|Z(G)|$ is divisible by 9, but I don't see how we might show that. Then it looks like $Z(G)$ containing a Sylow 3-subgroup would somehow imply that $Z(G) = G$ (and hence $G$ is abelian), but I'm not sure how that would work as well.

donburi
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1 Answers1

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EDIT: A user below points out a silly mistake I made. Make note of the changes.

This is the old trick that I mentioned here.

You get that $G$ acts on $P$ (where $P$ is your Sylow $3$-subgroup) by conjugation giving you a group map $G\to\text{Aut}(P)$ with kernel $C_G(P)$. Now, you know that $P$ is either $\mathbb{Z}_{9}$ or $\mathbb{Z}_3^2$. In the former case you have that $|\text{Aut}(P)|=|(\mathbb{Z}_9)^\times|=\varphi(9)=6$ and in the latter case you have that $|\text{Aut}(P)|=(3^2-1)(3^2-3)=48$ (this was computed because $\text{Aut}(\mathbb{Z}_3^2)\cong\text{GL}_2(\mathbb{F}_3)$ which has order $48$ by counting the number of vectors each column could be--they have to be lin. ind.). Either way, let's prove that $G/C_G(P)=\{e\}$.

The second case is easier. Namely, $|G/C_G(P)|$ divides both $|G|=315$ and $|\text{Aut}(P)|=48$. But, notice that since $P\leqslant C_G(P)$ one has that $9\mid C_G(P)$ and so $|G/C_G(P)|\mid 5\cdot 7$. So then, $|G/C_G(P)|\mid (5\cdot 7,48)=1$.

In the former case, you have that $|G/C_G(P)|$ divides both $|\text{Aut}(\mathbb{Z}_9)|=6$ and $G/C_G(P)$. But, you know that $P\subseteq C_G(P)$ so that $|G/C_G(P)|\mid 5\cdot 7$. So, you see that $|G/C_G(P)|$ divides both $35$ and $6$. Thus, once again, we see that $|G/C_G(P)|=1$ and so $G=C_G(P)$.

Either way we see that $G=C_G(P)$ and so $P\leqslant Z(G)$. Once again, we make the observation that $G/P$ is order $35$ which is cyclic. And, using the fact that if a quotient of $G$ by a subgroup of its center is cyclic, then $G$ is abelian, we may conclude that $G$ is abelian.

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Alex Youcis
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  • This is a nice proof, and I learned a few new ideas from your previous post (and your related blog post as well). However, I'm just wondering if there is a more elementary approach to the exercise as this approach uses quite a few theorems and concepts we didn't cover in class. – donburi Dec 12 '12 at 06:30
  • The automorphism group of $\mathbb{Z}/9\mathbb{Z}$ is $(\mathbb{Z}/9\mathbb{Z})^\times$ and has order 6, not 8. – Alex B. Dec 12 '12 at 13:13
  • Hi Alex. You write "The second case is easier. Namely, $|G/C_G(P)|$ divides both $|G|=315$ and $|\text{Aut}(P)|=48$ which implies that $|G/C_G(P)|=1$". But this doesn't imply that $|G/C_G(P)|=1$. It could be that $|G/C_G(P)|=3$. Secondly, you have a typo: it should say $|\text{Aut}(P)|=|(\mathbb{Z}_9)^\times|$. – Prism Aug 16 '13 at 19:24
  • @Prism I'm sorry. You can make the observation again that $P\leqslant C_G(P)$ so that $9\mid |C_G(P)|$ so that $3\nmid |G/C_G(P)|$. Does that make sense? – Alex Youcis Aug 19 '13 at 20:52
  • Yes it makes sense :) I just thought it should be clarified. Thanks for this great answer, and technique! (+1) – Prism Aug 19 '13 at 20:57
  • @Prism You're welcome! Thanks for pointing out the typos :) – Alex Youcis Aug 19 '13 at 20:59
  • Here's a different solution: $G/P$ is cyclic, so it has a normal subgroup of order $5$. Thus $G$ has a normal subgroup of order $3^2 \cdot 5$, which has a normal $5$-Sylow. Thus $G$ also has a normal $5$-Sylow. Similarly $G/P$ has a normal subgroup of order $7$ so $G$ has a normal subgroup of order $3^2 \cdot 7$ and thus a normal $7$-Sylow. Since each Sylow subgroup of $G$ is normal and abelian, $G$ is abelian. – Mikko Korhonen Aug 19 '13 at 21:30