Group of order $p^2$ is commutative with prime $p$

Question

Please help me on this one:

Let $p$ be a prime number, show that each group of order $p^2$ is commutative.

If you do not mind at all, could you please not give me the elegant explanation, but instead only the dummies' way? I know it must be against your craft skill, but I am self-studying and my knowledge is spotty at the very best. Thank you for your patience.

PS. The text then hints using these lemmas, but if you have simpler explanation, do please give me that one instead. Thanks again.
Lemma 1: If $G$ finite and $G/Z(G)$ cyclic, $G$ is commutative.
Lemma 2: [W. Burnside] $G$ is called primary if there exists a prime number $p$ such that $G$ is a $p$-group. If $G \neq \{1\},$ then $Z(G) \neq \{1\}.$

Answer 1

The center $Z(G)$ is non trivial. If $|Z(G)|=p^2$ then it is obvious.

If $|Z(G)| = p$ then $|G/Z(G)| = p$ and so $G/Z(G)$ is cyclic because every group of order $p$ with $p$ prime is cyclic. This implies that $G$ is abelian.

Infact Suppose $G/Z(G) = \langle{\overline{g}}\rangle$ and $a, b \in G$.

Thus in $G/Z(G)$ we have$$\overline{a} =\overline{g}^h $$ $$ \overline{b} =\overline{g}^k$$ for some $h, k \in \mathbb{N}$. So by definition of lateral class there exist $z_1 , z_2 \in Z(G)$ with

$$a = g^h z_1 \ \ \ \ b = g^k z_2$$

Thus $$ab = g^h z_1g^k z_2 = g^hg^k z_1z_2 = g^{h+k}z_2 z_1 = g^kg^h z_2 z_1 = g^k z_2g^h z_1 = ba$$i.e $G$ is abelian.

Answer 2

Theorem :(first sylow theorem) suppose that $G$ is a finite group and $|G|=p^{n}m$ such that $p \nmid m$ then

1.$G$ include a subgroup of order $p^i$ for $1\leq i \leq n$ 2. Any subgroup $H$ from $G$ from order $p^i$ is normal subgroup of a subgroup from order $p^i$

Answer : if $G$ be not cyclic then for every elemen except $e$ must have order $p$. Let $|\langle{a}\rangle|=p$ then $\langle{a}\rangle \neq G$,then there exist $b \in G$ such that $$|\langle{b}\rangle|=p \langle{a}\rangle \cap\langle{a}\rangle = \{e\}$$

By theorem $\langle{a}\rangle$ is normal in subgroup of order $p^2$ in fact it is normal in $G$. Also $\langle{b}\rangle$ is normal in subgroup of $G$ then we claim that

$$G \cong \langle{a}\rangle \times \langle{b}\rangle$$

So $G$ is commutative

Answer 3

Suppose $G$ is nonabelian. Conjugation induces a map from $G \to \operatorname{Aut}(G)$ that descends to a map $f:G/Z(G) \to \operatorname{Aut}(G/Z(G))$. Since finite $p$-groups have nontrivial center, $Z(G) = \mathbb{Z}_p$. Hence $f:\mathbb{Z}_p \to \mathbb{Z}_p^\times$, and so must be trivial. The result follows.