# There exists only two groups of order $p^2$ up to isomorphism.

I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.

Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.

Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?

## 4 Answers

If the (already known to be abelian) group $H$ in question is not cyclic, pick any nonzero element $a$. Its order must be $p$, so $\langle a\rangle$ is a subgroup of order $p$. Finally, consider $H/\langle a\rangle$, or, alternatively, pick another arbitrary element $b$ from $H\setminus\langle a\rangle$, and aim to prove that $\langle a\rangle \cap \langle b\rangle=\{ e\}$ and $\langle a,b\rangle=H$.

• Ah, yes. Sorry. – Berci Mar 1 '13 at 0:10
• Yes, your last edit is what I had in mind! – Pedro Tamaroff Mar 1 '13 at 0:16
• Surely the subgroups must intersect at the identity... – Potato Mar 1 '13 at 0:47
• I have taken the liberty of making this small correction. I hope you don't mind. – Potato Mar 1 '13 at 0:48
• @Potato Oddly enough, I guess most people read that correctly. At least, I did. – Pedro Tamaroff Mar 1 '13 at 0:57

Let $G$ be a $p^2$ group. As you said, it is Abelian.

Note that the order of every element divides $p^2$, so is equal to $1$ (for the identity $e$ only), $p$, or $p^2$.

If there is an element $x$ of order $p^2$, then $G=\langle x\rangle$ by cardinality. So $G$ is cyclic, and as you pointed out $$G\simeq \mathbb{Z}/p^2\mathbb{Z}.$$

Now assume that there is no element of order $p^2$. This means that every element which is not the identity has order $p$. Pick $x$ order $p$. Since $\langle x \rangle\subsetneq G$, you can take another order $p$ element $y$ in the complement of $\langle x \rangle$.

Now $$\theta:(u,v)\longmapsto uv$$ yields a homomorphism from $\langle x \rangle\times\langle y \rangle$ to $G$. Note that $\langle x\rangle\cap\langle y\rangle=\{e\}$, so the latter is injective. Since both groups have the same cardinality $p^2$, it follows that $\theta$ is an isomorphism.

Finally, since $\langle x\rangle \simeq\langle y\rangle \simeq \mathbb{Z}/p\mathbb{Z}$, we have $$G\simeq \langle x \rangle\times \langle y \rangle\simeq \mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}.$$

So $G$ is either isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$.

• Note that \langle and \rangle are much more convenient to use than >,<. – Pedro Tamaroff Mar 1 '13 at 0:34
• Thanks for the edit. I'm lazy sometimes... – Julien Mar 1 '13 at 0:34

The general fact that is useful here is the following:

Let $G$ be a group and $H,K$ subgroups of $G$ such that

1. $H\cap K=\{1\}$
2. $HK=G$
3. $H,K\unlhd G$

Then $G\cong H\times K$

Now, if $G$, your group of order $p^2$, is not $\mathbb{Z}_{p^2}$ then there exists $a,b\in G$ such that $\langle a\rangle\cap\langle b\rangle=\{1\}$ and $|a|=|b|=p$. Since

$$\left|\langle a\rangle\langle b\rangle\right|=\frac{|\langle a\rangle||\langle b\rangle|}{|\langle a\rangle\cap\langle b\rangle|}=p^2$$

you know that $\langle a\rangle\langle b\rangle=G$ and so from the above fact $G\cong \langle a\rangle\times\langle b\rangle\cong \mathbb{Z}_p^2$.

• One being normal gives us only a semi direct, not necessarily direct, product. – Alex Youcis Mar 2 '13 at 17:11
• Nevermind, I see where I was mistaken. – Pedro Tamaroff Mar 2 '13 at 17:25
• what means 3. ? – reuns Apr 2 '16 at 9:17

A proof of that could be:

The center of a group is a subgroup, so it's order must divide $p^2$, but it's a known fact that if a group has order $p^m$, with $p$ prime, then the center of the group is different from $p^{m-1}$ and different from $1$, so in our case, the center has order $p^2$ So it's abelian.

That's what you already have, know, by the theorem of structure of finite abelian groups, the group can bez either: $\mathbb{Z}_p\times\mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$. But from that same theorem, one can deduce that a group $\mathbb{Z}_m\times\mathbb{Z}_n$ is isomorphic to a group $\mathbb{Z}_{nm}$ iff $\gcd(m,n)=1$, so in our case, those two groups are not isomorphic, and there are only two groups of order $p^2$.

• I see nowhere in your answer that you address my question, except in the last sentence where you just enunciate what I aim to prove. – Pedro Tamaroff Mar 1 '13 at 0:18
• @PeterTamaroff But I proved it! I proved they're not isomorphic so they're different. From prooving they're all abelian and there's only two different, then there exist only two groups of that order. – MyUserIsThis Mar 1 '13 at 0:18
• I had already shown they weren't isomorphic. Why does "From proving they're all abelian and there's only two different, then there exist only two groups of that order." hold? – Pedro Tamaroff Mar 1 '13 at 0:22
• @PeterTamaroff You want to prove that there are only two groups of that order: you first prove that if a group has that order, it's abelian, and then that there are only two abelian groups of that order, so if a group has that order, it must be one of those. You're done. What else are you trying to prove? – MyUserIsThis Mar 1 '13 at 0:45
• @MyUserIsThis He presumably wants a more elementary proof that doesn't appeal to the classification of finite abelian groups. – Potato Mar 1 '13 at 0:46