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What is group isomorphic to $\text{Aut}(\mathbb{Z}_4 \times \mathbb{Z}_8)$?
I don't even know order of this group.
Please give me hint or proof outline.

Thanks!

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  • $\begingroup$ You can explore this group with GAP. Construct it with g:=AutomorphismGroup(DirectProduct(CyclicGroup(4),CyclicGroup(8)));. Now Size(g) returns 128 and StructureDescription(g) returns "(C2 x (((C4 x C2) : C2) : C2)) : C2" where : denotes a semidirect product. $\endgroup$ – Alexander Konovalov Mar 29 '17 at 13:05
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    $\begingroup$ Solution goes like in this question. $\endgroup$ – Dietrich Burde Mar 29 '17 at 13:11
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Let $m\neq n$, and $p$ be a prime. Then order of the group ${\rm Aut}(Z_{p^m} \times Z_{p^n})$ is given by $$ \phi(p^m)\phi(p^n)p^{2\min(m,n)}. $$ This has been shown at MSE here. In our case we have $p=2$ and $(m,n)=(2,3)$, so that the order is $128$. This coincides with the result using GAP.

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