Proof 1: Let \(G=\Z_{p^2}\times \Z_{p^2}\) and consider the subgroup \(H=\langle p \rangle\times \langle p \rangle \text{.}\) Then \(|H|=p^2\) since \(|p|=p\) in \(\Z_{p^2}\text{.}\) So \(H\) has \(p^2\) cosets in \(G\) as \(|G|=p^4\text{.}\) Define \(\phi: G/H \to \Z_p\times \Z_p\) by
\begin{equation*}
\phi((a,b)+H)=(a\bmod p,b\bmod p).
\end{equation*}
-
\(\phi\) is well-defined: Suppose \((a,b)+H=(c,d)+H\text{.}\) Then \((a,b)=(c,d)+(k_1p,k_2p)\) for some integers \(0\leq k_1,k_2\lt p\text{.}\) Thus we have \(a=c+k_1p\) and \(b=d+k_2p\text{.}\) So
\begin{align*}
\phi((a,b)+H)&=(a\bmod p,b\bmod p)\\
&=((c+k_{1}p)\bmod p,(d+k_{2}p)\bmod p)\\
&=(c\bmod p,d\bmod p)=\phi((c,d)+H)\text{.}
\end{align*}
-
\(\phi\) is surjective: Let
\((x,y)\in \Z_p\times \Z_p\text{.}\) Then we have
\(\phi((x,y)+H)=(x,y)\) (as
\(x,y\in \Z_p\) implies
\(x,y\in \Z_{p^2}\) also). Note that the reason this argument works is that
\(\phi\) is well-defined (else
\((x,y)+H\) and
\((x',y')+H\) may be the same coset that this argument requires to map to distinct elements).
-
\(\phi\) is injective: Suppose
\(\phi((a,b)+H)=\phi((c,d)+H)\text{,}\) i.e.
\((a\bmod p,b\bmod p)=(c\bmod p,d\bmod p)\text{.}\) Then we have
\(a-c=0 \bmod p\) and
\(b-d=0\bmod p\text{,}\) so we can write
\(a=c+k_1p\) and
\(b=d+k_2p\text{.}\) But then
\((a,b)=(c,d)+(k_1p,k_2p)\) and
\((k_1p,k_2p)\in H\text{,}\) so
\((a,b)+H=(c,d)+H\text{.}\)
-
\(\phi\) is a homomorphism: For any \((a,b)+H, (c,d)+H \in G/H\text{,}\)
\begin{align*}
\phi((a,b)+H+(c,d)+H)&=\phi((a+c,b+d)+H)\\
&=((a+c)\bmod p,(b+d)\bmod p)\\
&=(a\bmod p,b\bmod p)+(c\bmod p,d\bmod p) \text{ (since } p \text{ divides } p^2 \text{)} \\
&=\phi((a,b)+H)+\phi((c,d)+H)\text{.}
\end{align*}
Therefore \(G/H\) is isomorphic to \(\Z_p\times \Z_p\text{.}\)
Proof 2: Let \(G=\Z_{p^2}\times \Z_{p^2}\) and define \(\phi: G \to \Z_p \times \Z_p\) by
\begin{equation*}
\phi(a,b)=(a\bmod p,b\bmod p).
\end{equation*}
We see that \(\phi\) is surjective since for any \((x,y)\in \Z_p\times \Z_p\) we have \(\phi(x,y)=(x,y)\) (as \(x,y\in \Z_p\) implies \(x,y\in \Z_{p^2}\) also). We also have that for any \((a,b), (c,d) \in G\text{,}\)
\begin{align*}
\phi((a,b)+(c,d))&=\phi((a+c,b+d))\\
&=((a+c)\bmod p,(b+d)\bmod p)\\
&=(a\bmod p,b\bmod p)+(c\bmod p,d\bmod p) \text{ (since } p \text{ divides } p^2 \text{)} \\
&=\phi((a,b))+\phi((c,d))\text{.}
\end{align*}
So \(\phi\) is a group homomorphism. It follows from the First Isomorphism Theorem that the desired subgroup \(H\) is \(\ker \phi\text{.}\) So we will finish by determining \(\ker \phi\text{.}\) Suppose \((a,b)\in \ker \phi\text{.}\) Then we have \((a\bmod p,b\bmod p)=(0,0)\text{,}\) so \(a\) is a multiple of \(p\) and \(b\) is a multiple of \(p\text{.}\) The set of all elements in \(\Z_{p^2}\times \Z_{p^2}\) of this form is exactly \(H=\langle p\rangle \times \langle p\rangle\text{.}\)
