TITLE: Automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$
QUESTION [6 upvotes]: How to determine the automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$ where $p$ is a prime? Or more specifically, how to determine the element of order $2$ in this group?

I got stuck here, since I only know that if two finite groups $H$ and $K$, where $(|H|,|K|)=1$, then $\text{Aut}(H\times K) = \text{Aut}(H) \times \text{Aut}(K)$. But $p$ and $p$ are not relatively prime.
Any hints or solutions are welcomed, thanks!

REPLY [4 votes]: As suggested by vadim123, the automorphism group is isomorphic to $\operatorname{GL}(2, p)$, the group of $2 \times 2$ invertible matrices over $\Bbb{Z}$.
If $p = 2$, the group $\operatorname{GL}(2, 2)$ is isomorphic to $S_{3}$, so you should be able to determine the involutions, which are
$$
\begin{bmatrix}0&1\\1&0\end{bmatrix}, \quad
\begin{bmatrix}1&1\\0&1\end{bmatrix}, \quad
\begin{bmatrix}1&0\\1&1\end{bmatrix}.
$$
If $p > 2$, you will have
$$
c = \begin{bmatrix}-1&0\\0&-1\end{bmatrix},
$$
a central element, and then the conjugacy class of
$$
b = \begin{bmatrix}1&0\\0&-1\end{bmatrix},
$$
which has order
$$
\frac{\lvert \operatorname{GL}(2, p) \rvert}{\lvert C_{\operatorname{GL}(2, p)}(b) \rvert}
=
\frac{(p^{2} - 1)(p^{2} - p)}{(p-1)^{2}}
=
(p+1)p.
$$