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This question already has an answer here:

Let's say $\mathbb{Z}_n$ is the cyclic group. and I know that $Aut(\mathbb{Z}_n)\cong\mathbb{Z}_n^{\times}$, the multiplicative group of $\mathbb{Z}_n$. so the question is, is there a way to express $\mathbb{Z}_n^{\times}$ as a direct sum of cyclic groups?

I just thought that since the order of $\mathbb{Z}_n^{\times}$ is $\phi(n)$, and since it is abelian, maybe I should use the divisors of $\phi(n)=d_1^{s_1}...d_k^{s_k}$ like $\mathbb{Z}_{d_1^{s_1}}\oplus...\oplus\mathbb{Z}_{d_k^{s_k}}$, but I realized that there are too many non-isomorphic abelian groups with same order: for example, $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus\mathbb{Z}_2$ are both abelian and of order $4$ but they are not isomorphic.

so, yes, I guess there is a way since it is abelian, but I don't know how to do it specifically. can you give me a hint?

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marked as duplicate by user228113, Frits Veerman, Claude Leibovici, user91500, Namaste group-theory Jun 14 '17 at 12:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Yes there is! To see this we need two results:

  • The multiplicative group of $\mathbb Z_2^n$ is isomorphic to $\mathbb Z_2\oplus\mathbb Z_2^{n-2}$ for $n\geq 3$ and the mutiplicative group of $\mathbb Z_{p^n}$ is cyclic for $p$ an odd prime.
  • The chinese remainder theorem, which tells us $\mathbb Z_{p_1^{\alpha_1}\dots p_n^{\alpha_n}}^\times\equiv \mathbb Z_{p_1^{\alpha_1}}^\times\oplus\dots \oplus \mathbb Z_{p_n^{\alpha_n}}^\times$
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  • $\begingroup$ Here is a proof of the first result: math.stackexchange.com/a/1295687/33907 $\endgroup$ – Jorge Fernández Hidalgo Jun 13 '17 at 23:28
  • $\begingroup$ the chinese remainder theorem wasn't like this: $\mathbb{Z}_{p_1^{a_1}...p_k^{a_k}}\cong\mathbb{Z}_{p_1^{a_1}}\oplus...\oplus\mathbb{Z}_{p_k^{a^k}}$? does this work for the multiplicative group of $\mathbb{Z}_n$? $\endgroup$ – user159234 Jun 13 '17 at 23:40
  • $\begingroup$ yes, it works because the multiplicative group of a a product of rings is the product of all the multiplicative groups $\endgroup$ – Jorge Fernández Hidalgo Jun 13 '17 at 23:41
  • $\begingroup$ why? and how does that fact imply it? $\endgroup$ – user159234 Jun 13 '17 at 23:43
  • $\begingroup$ $(a_1,a_2,\dots, a_n)$ is invertible if and only if each $a_i$ is invertible. $\endgroup$ – Jorge Fernández Hidalgo Jun 13 '17 at 23:51

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