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$\begingroup$

I've tried, but I can't solve the question. Please help me prove that:

$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.

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    Since this isn't a complete answer, I'm answering as a comment. First: find an injection from $\textrm{Aut}(\mathbb{Z}^n)$ into $U_n$. Then consider why every element of $U_n$ acts on $\mathbb{Z}^n$ in a natural way.2012-08-24

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