# Expressing the automorphism group of a cyclic group as a direct sum of cyclic groups [duplicate]

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?

## marked as duplicate by user228113, Frits Veerman, Claude Leibovici, user91500, Namaste group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 14 '17 at 12:59

• 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 $\mathbb Z_{p_1^{\alpha_1}\dots p_n^{\alpha_n}}^\times\equiv \mathbb Z_{p_1^{\alpha_1}}^\times\oplus\dots \oplus \mathbb Z_{p_n^{\alpha_n}}^\times$
• 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 $\mathbb{Z}_n$? – user159234 Jun 13 '17 at 23:40
• $(a_1,a_2,\dots, a_n)$ is invertible if and only if each $a_i$ is invertible. – Jorge Fernández Hidalgo Jun 13 '17 at 23:51