Finding a subgroup $H$ of $\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}}$ with given condition.

Question

Question: Find a subgroup $H$ of $\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}}$ such that $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H$ is isomorphic to $\mathbb{Z_p}\oplus\mathbb{Z_p}$

My attempt: it is given in Hint that, consider $H=\{x\in\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}} : x^p=(0,0)\}$

I first proved that $H$ is subgroup of $\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}}$ of order $p^2$. Then as $|(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H|=|\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}}|/|H|=p^2$

Hence we have $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H≈ \mathbb{Z_{p^2}}$ or $≈\mathbb{Z_p}\oplus\mathbb{Z_p}$

Now to show $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H≈\mathbb{Z_p}\oplus\mathbb{Z_p}$ we must show that $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H$ has no element of order $p^2$ but I think it has element of order $p^2$ so that $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H≈\mathbb{Z_{p^2}}$ hence subgroup $H$ given in hint is wrong! ** Or I had done some mistake. Please tell me how to show $(\mathbb{Z_{p^2}}\oplus\mathbb{Z_{p^2}})/H≈\mathbb{Z_p}\oplus\mathbb{Z_p}$

Please help...

Answer 1

Hint: From your observations above, it suffices to show that there is no element in $\mathbb{Z}_p^2\oplus \mathbb{Z}_p^2/H$ of oerder $p^2$. An element $x+H$ has order $p$ in $\mathbb{Z}_p^2\oplus \mathbb{Z}_p^2/H$ if $x^p\in H$, now consider the definition of $H$.