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I would like to show that if $G = \langle a \rangle$ with order $n$, then $\text{Aut}(G) \cong U(n)$.

I know that $\text{Aut}(G)$ is the group of bijections $\phi:G \rightarrow G$ under composition, but I don't know where to start.

Any help would be much appreciated.

balddraz
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Tim
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1 Answers1

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Note that, upto isomorphism, $G \simeq \Bbb Z_n$. So the idea is :

  • Define $T: \text{Aut}(\Bbb Z_n) \to U(n)$ by $T(f)=f(1)$
  • Prove, this $T$ is a required isomorphism!

Note that $f(1)$ is again a generator for $\Bbb Z_n$, so $\gcd(f(1),n)=1$. Hence $f(1) \in U(n)$ is meaningfull

Chinnapparaj R
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