Let $G$ be a non-abelian $p$-group of the order $|G|=p^5$ such that Frattini Subgroup $\Phi(G)$, Commutator Subgroup $G'$ and Center $Z(G)$ of $G$ are equal and Rank($G$)=$3$, Exponent($G$)=$p$ (there exist unique such group but i do not know why?).
I wish to find a normal subgroup $N$ of $G$ of the order $p^2$ other than $Z(G)$.
My try: Clearly $G=\langle x,y,z\rangle$ for some $x,y,z\in G$. Atleast one of the subgroups $H_1=\langle x,y\rangle,H_2=\langle y,z\rangle$ and $H_3=\langle x,z\rangle$ is not of the order $p^2$. Suppose $|H_1|\neq p^2$, then $|H_1|=p^3$ or $p^4$ and $H_1$ contains the center $Z(G)$.
Now I am unable to proceed.
Let $G = \langle x,y,z \rangle$. Since $Z(G)= [G,G]$ is generated by the three commutators $[x,y]$, $[x,z]$ and $[y,z]$, we may assume that $Z(G)=\langle t,u \rangle$ with $[x,y]=t$, $[x,z]=u$.
If $[y,z]=1$, then take $N= \langle y,t \rangle$.
If $[y,z] = t^iu^j \ne 1$, then $[y,z] = [x,y^iz^j]$ and we can take $N = \langle y^iz^j, t^iu^j \rangle$.