Group Theory

1. Nov 13, 2013

Lee33

1. The problem statement, all variables and given/known data

If $|G| = p^2$ where $p$ is a prime, then $G$ is abelian.

2. The attempt at a solution

The book my proof gives is confusing at the very last part.

Suppose $|Z(G)| = p$. Let $a\in G$, $a\notin Z(G)$. Thus $|N(a)| > p$, yet by Lagrange $|N(a)| \ | \ |G| = p^2$.

Why is $|N(a)| \ | \ |G| = p^2$? Since $|G| = p^2$ and $|N(a)|> p$ how does that make $|N(a)| \ | \ |G| = p^2$?

2. Nov 13, 2013

Dick

You are reading it wrong. |N(a)| | |G| just means that the number |N(a)| divides the number |G|. It doesn't have a numerical value. They are just saying |N(a)|>p and yet |N(a)| must divide |G|=p^2. So?

3. Nov 13, 2013

Lee33

So, since $|N(a)|>p$ and $|N(a)|$ must divide $|G|$ then $|N(a)| = p^2$?

4. Nov 13, 2013

Dick

Yes!

5. Nov 13, 2013

Lee33

Thanks! Last question, since $|N(a)| = p^2$ why would this imply $a\in Z(G)$?

6. Nov 13, 2013

Dick

N(a) is the normalizer subgroup of a. If N(a)=p^2 then N(a)=G, right? You tell me why that should mean $a\in Z(G)$.

7. Nov 13, 2013

Lee33

Since N(a) = G then N(a) generates the group G thus $a\in Z(G)$?

8. Nov 13, 2013

Dick

I don't think that really means anything. What does N(a) mean?

9. Nov 13, 2013

Lee33

It is the normalizer, so $N(a) = \{ x \in G \ | \ xa = ax\}.$

10. Nov 13, 2013

Dick

So in words, N(a) is the set of all elements of G that commute with a. Now what does $a\in Z(G)$ mean in words describing what a commutes with?

11. Nov 13, 2013

Lee33

The center of the group G is the set of elements that commute with every element of G. Where N(a) is the set of all elements of G that commute with a. So is it because then that will make G abelian hence G = Z(G) hence a in Z(G)?

12. Nov 13, 2013

Dick

It is going to be shown abelian eventually, but you aren't there yet, so you can't use 'abelian' to prove a is in Z(G). First just tell me why a is in Z(G). Think about the words and use that N(a)=G (why do we know that?).

Last edited: Nov 13, 2013
13. Nov 13, 2013

Lee33

Since N(a) = G then all elements of G commute with a thus why $a \in Z(G)?$

14. Nov 13, 2013

Dick

I'd feel more confident you understood that if you wouldn't end all of your comments in a '?'. If you remind me how we know N(a)=G that might restore my confidence. No '?', ok?

15. Nov 13, 2013

Lee33

Lol I understand. We got $N(a) = G$ since we establish that since $|N(a)|>p$ and $|N(a)|$ must divide $|G|$ then $|N(a)| = p^2$ since $p^2 = |G|$ then $|N(a)| = |G|$.

16. Nov 13, 2013

Dick

Right! N(a) is a subgroup of G having p^2 elements. But G has only p^2 elements. So |N(a)|=|G| means N(a) must equal G, which should be your final conclusion. Thinking about these things in words and using less symbols often helps.

17. Nov 13, 2013

Lee33

Yes, it does! Thank you very much!