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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.