# Automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$

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 $$\Aut(H\times K) = \Aut(H) \times \Aut(K)$$. But $$p$$ and $$p$$ are not relatively prime.

Any hints or solutions are welcomed, thanks!

## 2 Answers

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 $$\frac{\lvert \operatorname{GL}(2, p) \rvert}{\lvert C_{\operatorname{GL}(2, p)}(b) \rvert} = \frac{(p^{2} - 1)(p^{2} - p)}{(p-1)^{2}} = (p+1)p.$$

• Thanks! How to prove that all the element of order two is the conjugacy of $b$ or the central element? – Golbez May 25 '13 at 18:41
• @Golbez, let $p > 2$. An element $g$ of order $2$ is a root of the polynomial $x^{2} - 1$. If its minimal polynomial is $x + 1$, then $g$ is the central element $c$. If the minimal polynomial is $x^{2} - 1$, then $g$ is diagonalizable, that is, is conjugate to $b$. – Andreas Caranti May 25 '13 at 18:52
• @Thanks for your patience! – Golbez May 25 '13 at 18:56
• @Golbez, you're welcome! – Andreas Caranti May 25 '13 at 18:57

An automorphism of a vector space is defined by its action on a basis. So, take the unique automorphism determined by

$(1,0)\rightarrow (0,1)$ and $(0,1)\rightarrow (1,0)$.

• Thanks! I have got your idea. But how to exclude other cases, such that $(0,1)\to (a,b),(1,0)\to (c,d)$ where $(a,b)$ and $(c,d)$ are two basis? – Golbez May 25 '13 at 18:26