# Prove, that group of order $p^2$ is abelian.

I know there is a proof using these theorems:

• The center of a finite p−group is non-trivial

• For any group G , $G/Z(G)$ is cyclic iff $G$ is abelian, or in otherwords: the quotient $G/Z(G)$ can never be non-trivial cyclic.

But is there a proof not using these theorems?

• The only proof that I can think of contains essentially the same arguments as you would use in proving the two theorems you mention, so it more efficient to prove these theorems independently. – Derek Holt Dec 4 '16 at 16:26

## 3 Answers

If there is an element of order $p^2$, it's cyclic and thus abelian. Suppose there is no element element of order $p^2$. Then, the order of of the elements of $G$ are either $1$ or $p$. Let $h_1,h_2\in G$ two elements of order $p$ s.t. $h_2\notin\left<h_1\right>$. Then, $\left<h_1,h_2\right>$ is of order $p^2$ and is s.t. $|\left<h_1,h_2\right>|\geq p+1$. Therefore, $|\left<h_1,h_2\right>|=p^2$, and thus $G=\left<h_1,h_2\right>$. Therefore, $G$ is abelian.

We can show that $\left<h_i\right>$ are normal in $G$. Then, $[G:H_i]=p$ and thus $G/H_i$ are cyclic, and thus abelian. Let consider $$\pi: G\longrightarrow G/H_i,$$ defined by $\pi(g)=gH_i$. Take an element of $[G,G]=\left<ghg^{-1}h^{-1}\mid g,h\in G\right>$. You have that $$\pi(ghg^{-1}h^{-1})=\pi(g)\pi(h)\pi(g^{-1})\pi(h^{-1})\underset{G/H_i\ cyclic}{=}H_i$$ and thus and thus $[G,G]\leq H_i$, and since $H_1\cap H_2=\{1\}$, we get $[G,G]=\{1\}$. Therefore $G=G/[G:G]$ is abelian.

• Why does $h_1 h_2 = h_2 h_1$ ? – lhf Dec 4 '16 at 13:45
• @lhf: I completed my proof. – Surb Dec 4 '16 at 14:00
• But you haven't shown that the $\langle h_i \rangle$ are normal in $G$. – Derek Holt Dec 4 '16 at 16:24
• @DerekHolt: I will not do all the work ;-) But I agree, it's not the easiest part. – Surb Dec 4 '16 at 17:30
• btw $k_4$ is abelian not cyclic – RAM_3R Feb 25 at 22:20

Now for something completely different (using representation theory over $$\mathbb{C}$$)...

Let $$G$$ be a group of order $$p^2$$. Let $$Irr(G)$$ be its set of complex irreducible characters. Note that the principal character $$1_G \in Irr(G)$$. Since $$|G|=p^2=\sum_{\chi \in Irr(G)}\chi(1)^2=1 + \sum_{\chi \in Irr(G)-\{1_G\}}\chi(1)^2$$, and for all $$\chi \in Irr(G)$$: $$\chi(1) \mid p^2$$, it follows that all $$\chi$$ must be linear, which is equivalent to $$G$$ being abelian.

Sorry to revive a very old thread -- and if someone better versed in netiquette says so I'll just delete this -- but I was looking for a nice low-level proof of this fact (which is what brought me to this page in the first place) and was delighted to find two fairly straightforward and low-level proofs of this nice fact. Here's the slightly simpler one (basically thieved off the 'net somewhere, but I forget where); the most sophisticated fact here is Fermat's Little Theorem. I thought I'd record it in case anybody else is ever interested.

Let $$G$$ be a non-cyclic group of order $$p^2$$ with identity element $$e$$, and let $$g,h\in G$$ of order $$p$$ such that $$\langle g\rangle\cap \langle h \rangle = \{e\}$$. Assume by way of contradiction that $$g$$ and $$h$$ do not commute. I claim that $$ghg^{-1}$$ cannot be a power of $$h^k$$ of $$h$$, for if it were, then one calculates that $$g^rhg^{-r} = h^{k^r}$$ for all $$r$$, and then $$g^{p-1}hg^{1-p} = h^{k^{p-1}}$$, which is equal to $$h$$ by Fermat. This would imply $$g^{-1}h=hg^{-1}$$ and now $$gh=hg$$, contradiction.

So $$\langle h\rangle$$ and $$\langle ghg^{-1}\rangle$$ are distinct subgroups of order $$p$$, and the distinct cosets of $$\langle ghg^{-1}\rangle$$ in $$G$$ are $$\langle ghg^{-1}\rangle, h\langle ghg^{-1}\rangle, h^2\langle ghg^{-1}\rangle, \dots, h^{p-1}\langle ghg^{-1}\rangle$$. Therefore $$g^{-1}$$ lies in some $$h^k\langle ghg^{-1}\rangle$$, so $$g^{-1}=h^kgh^{\ell}g^{-1}$$ for some $$\ell$$. Then $$e=h^kgh^{\ell}$$ and $$g=h^{-k-\ell}$$, contradicting that $$\langle g \rangle$$ and $$\langle h \rangle$$ intersect trivially.

Well, now we are done: $$g$$ and $$h$$ commute, so the mapping from $$\mathbb Z_p\times \mathbb Z_p$$ to $$G$$ taking $$(a,b)$$ to $$g^ah^b$$ is an isomorphism.