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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.

Myshkin
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3 Answers3

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You have almost everything that you need already. The last step is supplied by Euler’s totient function.

Brian M. Scott
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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.

MJD
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  • You also need to explain why taking a generator to any other generator produce an automorphism is this soulution. – Belgi Jul 22 '12 at 21:59
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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 $U_{144} \simeq \mathbb{Z}_6 \times \mathbb{Z}_8$.

Nicky Hekster
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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.

     – Mikko Korhonen Jul 26 '12 at 16:37
  • The automorphism group of $\mathbb{Z}{16}$ is not (cyclic) $\mathbb{Z}{8}$. Rather it's isomorphic to $\mathbb{Z}{2}\times\mathbb{Z}{4}$. – SPDR Jun 19 '25 at 17:28