How to prove that a Group $G$ with order $p^2$ must be a abelian group(where p is a prime)?

Question

I consider that the elements in $G$ only have two cases:

1) $G$ has an element $a$ with order $p^2$ then it's abelian.

2) $G$ has an element $a$ with order $p$, then we have a subgroup $<a>$ and choose another element $b$ $\notin <a>$ then we have $G = <a,b>$ but under this circumstance, how to prove that $<a,b>$ is abelian?

Edit: With the help of the comments posted by @user1729 and @JohnHughes, I may have a solution:

Consider $Z(G)$, since the order of $G$ is $p^2$ then we know $G$ has nontrivial center, also the center of $G$ is a normal subgroup of $G$, we get the followings results:

(1) $Z(G)$ is abelian with order prime $p$ (2) $G/Z(G)$ has order prime $p$ then is abelian.

Using(1) and (2) we know $G$ is abelian.

Edit: The last edit is wrong, since I do not use that $Z(G)$ is generated by a single element $a$ and $ab = ba$ then $G = <a,b>$ is abelian.

Do you know that (1) if $|G|=p^n$ for $p$ prime then $Z(G)\neq1$, and (2) if $G/Z(G)$ is cyclic then $G$ is abelian? If so, combine these fact to prove your result. — user1729, May 15 '18 at 13:25
Hint: Nothing you've done so far depends on $p$ being prime. (You also didn't mention that $p$ was prime in the question, but it's essential. For instance, there are groups of order $16$ that are not abelian.) — John Hughes, May 15 '18 at 13:25
OK, but you haven't actually used the prime-ness in any way. — John Hughes, May 15 '18 at 13:28
It is unclear how your edit helps. In particular, there are non-abelian groups $G$ such that $N$ and $G/N$ are both abelian (e.g. $S_3$). — user1729, May 15 '18 at 13:44

score 2 · Accepted Answer · answered May 15 '18 at 13:30

The key here is that the center is nontrivial. Once you have that, you can take $a$ belonging to the center, then $a$ commutes with $b$ so the group generated by $a$ and $b$ must be abelian.

So try to prove that the center of a group of order $p^2$ for $p$ prime is nontrivial.

How to prove that a Group $G$ with order $p^2$ must be a abelian group(where p is a prime)?

1 Answers1