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we have to find number of automorphism on $\mathbb{Z}_9\times \mathbb{Z}_{16}$

I know a result which says $Aut(\mathbb{Z}_n)\cong U_n$ where $U_n$ is the multiplicative group i.e $$U_n=\{x:(x,n)=1\}$$

I know another result, $\mathbb{Z}_n\times \mathbb{Z}_m\cong \mathbb{Z}_{mn}\Leftrightarrow(m,n)=1$,well,could any one give me Hint for this problem? thank you for help.

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    You have everything that you need right there.2012-07-22
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    @BrianM.Scott so it's $U_{9*16}$, right ?2012-07-22
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    @Belgi: Yes, $U_{144}$.2012-07-22
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    yes sir ah! its $U_{144}$2012-07-22
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    @J.D. done!....2012-07-22

3 Answers 3

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Observe that $Aut(\mathbb{Z}_m \times \mathbb{Z}_n)\simeq Aut(\mathbb{Z}_m) \times Aut(\mathbb{Z}_n)$, whenever $m$ and $n$ are relatively prime. This reflects by the way the multiplicativity of the Euler totient function. Hence $U_{144} \simeq \mathbb{Z}_6 \times \mathbb{Z}_8$.

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    In general, it is true that $$\operatorname{Aut}(G\times H) \cong \operatorname{Aut}(G) \times \operatorname{Aut}(H)$$ when $G$ and $H$ are finite groups of relatively prime order.2012-07-26
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You have almost everything that you need already. The last step is supplied by Euler’s totient function.

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Here's your hint: You know that ${\Bbb Z}_9\times {\Bbb Z}_{16} = {\Bbb Z}_{144}$. An automorphism of a cyclic group is completely determined by the image of the generator. So all you have to do is count how many elements of ${\Bbb Z}_{144}$ are generators.

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    You also need to explain why taking a generator to any other generator produce an automorphism is this soulution.2012-07-22