The Problem: $G$ noncyclic, $|G|=p^3$, $p$ an odd prime, . Let $E=\text{ker}(\varphi)$, where $\varphi: G\to Z(G)$ is $x\mapsto x^p$. Show that if $G$ contains an element of order $p^2$, then $E\cong Z_p\times Z_p$.
I am aware that there is a slight difference between my title and my problem; but note that "$|\text{ker}(\varphi)|=p^2$" implies "$G$ contains an element of order $p^2$".
Source: Abstract Algebra, $\mathit{3^{rd}}$ edition by Dummit and Foote.
In Section 5.5 (Semidirect Product), where the book discusses the classification of groups of order $p^3$, $p$ an odd prime, one paragraph goes:
Let $x$ be an element of order $p^2$ and let $H=\left\langle x\right\rangle$. Note that since $H$ has index $p$, $H$ is normal is $G$ by Corollary 4.5. If $E$ is the kernel of the $p^{th}$ power map, then in this case $\mathbf{E\cong Z_p\times Z_p}$ and $\mathbf{E\cap H=\left\langle x^p\right\rangle}$.
My Question: I struggle to show the boldface texts in the quoted paragraph. I understand that either $E\cong Z_{p^2}$ or $E\cong Z_p\times Z_p$; but I failed to deduce a contradiction by assuming the former. And even if we derived the latter (i.e., $E\cong Z_p\times Z_p$), how do we know that $E\cap H=\left\langle x^p\right\rangle$? Any hint would be greatly appreciated.
The kernel is made of $p^2-1$ elements of order $p$, plus the identity. So, $p^2-1=k(p-1)$, where $k$ is the number of subgroups of $E$ of order $p$, namely $k=p+1$. Take any two of them, say $K$ and $L$. Since $K\cap L=\{e\}$, then $|KL|=p^2=|E|$; but $KL\subseteq E$, then $KL=E$. Moreover, $E$ is abelian (as its order is $p^2$). So, $E\cong K\times L\cong Z_p\times Z_p$.