We prove every group $G$ of order $p^2$ is either cyclic or isomorphic to $\mathbb Z_p\times \mathbb Z_p$

first we prove it is abelian, this is an immediate consquence of these three lemmas:

lemma $1$: If $\frac{G}{Z(G)}$ is cyclic then $G$ is abelian.

lemma $2$: the center of a finite $p$-group is not trivial.

lemma $3$: a group of prime order is cyclic.

Once we know it is abelian there are two cases:

case $1$: $G$ has an element of order $p^2$, and hence is cyclic

case $2$: $G$ has no element of order $p^2$, hence by Lagrange's theorem all of its non identity elements have order $p$. take an element $g$ of order $p$ and then take a second element $h\in G\setminus \langle g \rangle $ with order $p$. We have $|\langle g \rangle \langle h \rangle |=\frac{|\langle g \rangle | |\langle h \rangle |}{|\langle g \rangle \cap \langle h \rangle | }$. Clearly the denominator is $1$ by lagrange's theorem, so these two subgroups are both normal, their intersection is trivial and their product is all of $G$. This implies $G\cong \langle p \rangle \times \langle q\rangle \cong \mathbb Z_p\times \mathbb Z_p$