Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
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 6:30
    
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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