Unraveling the Mystery of Abelian Groups When ##|G| = p^2##

[Lee33]

Homework Statement

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 $|N(a)| \ | \ |G| = p^2$?

 

[Dick]

Homework Statement

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 $|N(a)| \ | \ |G| = p^2$?

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?

 

[Lee33]

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

 

[Dick]

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

Yes!

 

[Lee33]

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

 

[Dick]

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

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 $a\in Z(G)$.

 

[Lee33]

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

 

[Dick]

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

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

 

[Lee33]

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

 

[Dick]

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

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?

 

[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)?

 

[Dick]

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)?

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?).

 

[Lee33]

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

 

[Dick]

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

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?

 

[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 $|N(a)| = |G|$.

 

[Dick]

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 $|N(a)| = |G|$.

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.

 

[Lee33]

Yes, it does! Thank you very much!