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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!

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2 Answers 2

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

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  • $\begingroup$ Thanks! How to prove that all the element of order two is the conjugacy of $b$ or the central element? $\endgroup$
    – Golbez
    May 25, 2013 at 18:41
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    $\begingroup$ @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 $b$. $\endgroup$
    – Andreas Caranti
    May 25, 2013 at 18:52
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    $\begingroup$ @Thanks for your patience! $\endgroup$
    – Golbez
    May 25, 2013 at 18:56
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    $\begingroup$ @Golbez, you're welcome! $\endgroup$
    – Andreas Caranti
    May 25, 2013 at 18:57
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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 $(0,1)\rightarrow (1,0)$.

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  • $\begingroup$ 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? $\endgroup$
    – Golbez
    May 25, 2013 at 18:26

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