I happen to know that Aut($\mathbb{Z_p} \times \mathbb{Z_p}$) $\simeq$ $GL_2(p)$, and so has order $p(p^2-1)(p-1)$.

Which means something is wrong with the following argument:

An automorphism $\phi$ is determined by its value on the two generating elements $(1,0) $ and $(0,1)$ of $\mathbb{Z_p} \times \mathbb{Z_p}$.

These must be sent to elements of the same order, $p$. In fact every nonzero element of $\mathbb{Z_p} \times \mathbb{Z_p}$ has order $p$, so there are $p^2-1$ choices to which each generator must go. Therefore the number of automorphisms is $(p^2-1)^2$.

So I've overcounted by $(p^2-1)(p-1)$; I imagine some of the homomorphisms in my argument were not distinct, or not isomorphisms. is there a way to easily rectify for the overcounting? Or do we throw the whole argument away?


You have indeed counted homomorphisms that are not isomorphisms.

You are correct in saying that there are $p^2-1$ choices for where one of the generators, say $(1,0)$, is mapped. However, once you have chosen that, you have also determined the images of $(2,0)$, $(3,0)$, $\dots$, $(p-1,0)$. You cannot map the second generator to the same thing as any of those elements listed, so this excludes $p-1$ elements. Of course we also want to exclude $0$, so that leaves $p^2-p$ choices for where the second generator is mapped.

This gives the correct order, $(p^2-1)(p^2-p)$.

  • $\begingroup$ Wonderful. Let's say I did not know a priori that this is the correct order. Is there any method that comes to your mind of proving that it is in fact the correct order (and not merely an upper bound)? That is to say, how trivial is it to claim that each of these homomorphisms are distinct, and isomorphisms? Thank you! $\endgroup$ – SSF Nov 9 '17 at 3:13
  • 1
    $\begingroup$ It's clear that each of them is distinct. To show they're isomorphisms you can try writing down an inverse, which will essentially be the formula for the inverse of a matrix. $\endgroup$ – Alex Zorn Nov 9 '17 at 3:16

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