# 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. Dec 4, 2016 at 16:26

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, 2016 at 13:45
• @lhf: I completed my proof.
– Surb
Dec 4, 2016 at 14:00
• But you haven't shown that the $\langle h_i \rangle$ are normal in $G$. Dec 4, 2016 at 16:24
• @DerekHolt: I will not do all the work ;-) But I agree, it's not the easiest part.
– Surb
Dec 4, 2016 at 17:30
• btw $k_4$ is abelian not cyclic Feb 25, 2019 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.

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.

Here I am referencing a solution given on Aryaman Maithani's github page. These steps can lead the proof straight forwardly and they are:

Firstly, use the first lemma you mentioned to conclude that $$Z(G)$$ is of order $$p$$ or $$p^2$$. The latter case is trivial by the definition of the center. In the former case, take an element $$x$$ from $$G\setminus Z(G)$$ and consider all the elements of $$G$$ that commute to $$x$$, name that set $$H$$. Prove that this set $$H$$ is a subgroup of $$G$$. Because $$x\in H$$ we have that $$Z(G)$$ is a proper subset of $$H$$. Then, $$H$$ is a proper subset of $$G$$ because $$x$$ isn't in $$Z(G)$$. Now the order of $$H$$ must be strictly between $$p$$ and $$p^2$$ and must divide $$p^2$$. This is obviously impossible and that is the contradiction for the former case.

We use the fact that the center of any $$p-$$group is non-trivial (this uses the class equation). Since the order of $$G$$ is $$p^2$$, by LaGrange's theorem as $$Z(G) \leq G$$, we have that either (as $$Z(G) \neq \{e\}$$) $$\vert Z(G) \vert=p$$ or $$p^2$$. If $$\vert Z(G) \vert = p^2$$, then $$Z(G)=G$$ and $$G$$ is abelian. Suppose then that $$\vert Z(G) \vert = p$$. Then $$Z(G) \varsubsetneq G$$ and $$G$$ is non-abelian. Then $$\vert G/ Z(G) \vert = \frac{p^2}{p}=p$$, then $$G/Z(G) \cong \Bbb{Z} / p \Bbb{Z}.$$ And we know $$\Bbb{Z} / p \Bbb{Z}$$ is cyclic hence $$G/Z(G)$$ is cyclic thus $$G$$ is abelian (fact that $$G$$ is abelian iff $$G/Z(G)$$ is cyclic), contradiction as $$G \neq Z(G)$$, thus $$Z(G)$$ has order $$p^2$$ hence $$Z(G)=G$$ forcing $$G$$ to be abelian.