There are exactly two abelian groups of order $p^2$

Question

Let $p$ be a prime number. up to isomorphism, there is exactly two abelian groups of order $p^2$.

It is easy to see those two are $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$. It is easy to see that if their exist another one then it should not be cyclic. But how to prove exactly?

Possible duplicate of There exists only two groups of order $p^2$ up to isomorphism.. — Dietrich Burde, Aug 28 '17 at 18:53

score 5 · Accepted Answer · 2017-08-28T19:34:32.000

5

Consider 2 cases: whether or not there exists an element of order $p^2$. If it does exist, then the group is $\mathbb{Z}_{p^2}$. If not, then you have to prove that this condition forces the group to be $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$

To prove that consider two elements $x \in G, y \in G\backslash\langle x \rangle$ of order $p$ and the group generated by them:

$| \langle x, y \rangle | > | \langle x \rangle |$, hence $G = \langle x, y \rangle.$ It is easy to see from here that $G \cong \mathbb{Z}_{p}\times\mathbb{Z}_{p}$

edited Aug 28 '17 at 19:34

answered Aug 28 '17 at 18:39

So every non-cyclic groups are isomorphic to $\mathbb{Z}_p\times \mathbb{Z}_p$, that's all, isn't? – MAN-MADE Aug 28 '17 at 19:20
@MANMAID Yes, it is – Aug 28 '17 at 19:33

score 0 · Answer 2 · answered Aug 28 '17 at 18:39

0

See here fundamental theorem of finitely generated abelian groups.

answered Aug 28 '17 at 18:39

Davood

4,265

1

This is way overkill. – Tobias Kildetoft Aug 28 '17 at 18:43
1

See also this duplicate, which is not using an "overkill". – Dietrich Burde Aug 28 '17 at 18:53

There are exactly two abelian groups of order $p^2$

2 Answers2