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I'd like to calculate the orders of the following automorphism groups where $p$ and $q$ are distinct primes:

(1) $\mathrm {Aut}(\mathbb Z_p \times \mathbb Z_p \times \mathbb Z_p \times \mathbb Z_p)$

(2) $\mathrm {Aut}(\mathbb Z_p \times \mathbb Z_p \times \mathbb Z_q)$

I figured out that if $m$ and $n$ are relatively prime, then $\mathrm {Aut}(\mathbb Z_m) \times \mathrm {Aut}(\mathbb Z_n)$ is isomorphic to $\mathrm {Aut}(\mathbb Z_m \times \mathbb Z_n )$, but I could'nt apply this one in this case.

I already solved the case (1). For (2), is $\mathrm {Aut}(\mathbb Z_p \times \mathbb Z_p \times \mathbb Z_q)$ isomorphic to $\mathrm {Aut}(\mathbb Z_p \times \mathbb Z_p ) \times \mathrm {Aut}( \mathbb Z_q)$? If not, what is a group isomorphic to $\mathrm {Aut}(\mathbb Z_p \times \mathbb Z_p \times \mathbb Z_q)$? And how can I compute the order?

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    If $G$ and $H$ are any two finite groups with coprime orders, then ${\rm Aut}(G \times H) \cong {\rm Aut}(G) \times {\rm Aut}(H)$. – 2017-02-05
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    Then can you provide the proof of that fact? – 2017-02-05
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    If $G$ and $H$ have coprime orders, then they are characteristic subgroups of $G \times H$. Or, in other words, under that assumption, $G$ consists of the set of elements of $G$ whose orders divide $|G|$, and so any automorphism of $G \times H$ must fix the set $G$ and hence induce an automorphism of $G$. Similarly for $H$. – 2017-02-05

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Note how an integer matrix acts on the canonical basis of $\mathbb{Z}^n$.

$Aut(\mathbb{Z}^n) \cong GL_n(\mathbb{Z})$ the group of $n \times n$ unimodular matrices (integer matrices whose inverse is an integer matrix), i.e. the integer matrices with determinant $\pm 1$ ($\pm 1$ being the only units in $\mathbb{Z}$).

For $Aut(\mathbb{Z}_p^n) \cong GL_n(\mathbb{Z}_p)$ it works the same way, because the adjugate matrix formula $M \text{Adj}(M) = \det(M)I $ works in any domain. So $GL_n(\mathbb{Z}_p)= \{ M \in \mathbb{Z}_p^{n \times n}, \det(M) \not \equiv 0 \bmod p\}$.

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    Thanks a lot. Then how about the second one? – 2017-02-04
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    @bellcircle try and see ? – 2017-02-04
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    Also, how can I calculate the order of $GL_4(\mathbb Z_p)$? – 2017-02-04
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    @bellcircle possibly looking at the [generators of $GL_n(\mathbb{Z})$](http://math.stackexchange.com/questions/2129331/find-a-finite-generating-set-for-gln-mathbbz?noredirect=1#comment4379033_2129331) and adapting to $GL_n(\mathbb{Z}_p)$ will show it as a semi-direct product of some cyclic groups. – 2017-02-04
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    Okay, I solved the first one, and it remains only the second one. – 2017-02-05
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    http://math.uchicago.edu/~may/REU2013/REUPapers/Sommer-Simpson.pdf This one has answer. – 2017-02-05
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    @bellcircle ok, it is page 7 : $\gamma : \mathbb{Z}_p^2 \to \mathbb{Z}_p^2$ is an automorphism iff $\gamma((1,0)) \ne 0, \gamma((0,1)) \ne 0$ and $\langle \gamma((1,0)) \rangle \cap \langle \gamma((0,1)) \rangle = \{0\}$. For choosing $\gamma((1,0))$ we have $p^2-1$ choices, and for choosing $\gamma((0,1))$ we have $p^2-p$ choices, so $|Aut(\mathbb{Z}_p^2)| = (p^2-1)(p^2-p)$. And we apply the same idea to $|Aut(\mathbb{Z}_p^n)| = (p^n-p^0)(p^n-p^1)\ldots(p^n-p^{n-1})$ – 2017-02-05
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    @bellcircle But it doesn't explain how $GL_n(\mathbb{Z}_p)$ is the semi-direct product of some cyclic groups, as $GL_n(\mathbb{Z})$ (page 12..) – 2017-02-05
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    It doesn't matter for me because it suffices to compute the order of $\mathbb Z_p^4$. – 2017-02-05