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Let $G$ be a cyclic group of order $m$. What is the order of $\text{Aut}(G)$?

I want to know the proof as well (elementary if possible). I would still accept the proof if one answers with $m = p$, a prime. Or on top of that, I would accept the answer with extra assumption: $q \equiv 1$ mod $p$ with another prime $p$.

Curious student
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user123454321
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2 Answers2

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Since an automorphism of $G$ should map a generator of $G$ to a generator of $G$ it's enough to know how many generators does $G$ have.

If $G=\{e,g,g^2,...,g^{m-1}\}$ then a $g^i$ generates G if and only if $\operatorname{gcd}(i,m)=1$.

$\lvert \operatorname{Aut}(G)\rvert=\phi(m)$ where $\phi(m)$ is Euler's function.

For a more detailed proof:

  1. Let $G=\langle g\rangle$ and $f\in\operatorname{Aut}(G)$.
  2. Then $f(g)=g^i$ for some $i$. If $f$ is an isomorphism $\langle g^i\rangle =G$ and this happens only if $\operatorname{gcd}(i,m)=1$.
  3. On the other hand every homomorphism $f:G\rightarrow G$ with $f(g)=g^i$ is an isomorphism when $\operatorname{gcd}(i,m)=1$, so $\lvert \operatorname{Aut}(G)\rvert=\phi(m)$.
epsilon
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    In fact, the automorphism group is isomorphic with the multiplicative group of integers modulo $m$ (by a very similar argument). – tomasz Dec 30 '12 at 22:42
  • I totally agree tomasz. – epsilon Dec 30 '12 at 23:03
  • Thanks for editing tomasz , it looks much nicer now. – epsilon Dec 31 '12 at 00:06
  • Sometimes thinking in Greek and writing in English doesn’t produce a nice outcome lol :) – epsilon Dec 31 '12 at 00:07
  • You're welcome. :) – tomasz Dec 31 '12 at 01:52
  • How would you fill in the details of the third part of your proof? Why is it that if $f:G \to G$ with $f(g) = g^i$ is an isomorphism when $gcd(i,m)=1$? – objectivesea Apr 11 '13 at 21:24
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    @objectivesea Since $gcd(i,m)=1$ there exist $x_1,x_2$ such that $ix_1+mx_2=1$. It's easy to prove that s is an epimorphism if you see that $g^{ix_1+mx_2}=g\rightarrow g^{i x_1}=g$ since $g^{mx_2}=1$. To prove that it's a monomorphism suppose $g^{im_1}=g^{im_2}$ then $g^{i(m_1-m_2)}=g$ and $m|i(m_1-m_2)$. Since $gcd(m,i)=1\rightarrow m|m_1-m_2\rightarrow m_1=m_2+km$. $g^{im_1}=g^{im_2+ikm}=g^{im_2}$. That's it more or less. – epsilon Apr 12 '13 at 07:48
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Hint:

For a given cyclic group $G,\;\text{with}\;\; |G| = m$: $$\text{Aut}\,(G) \cong \mathbb{Z}_m^*\tag{$\;^*:\;\;$multiplicative group}$$

Hence $\text|{Aut}\,(G)| = |\mathbb{Z}_m^*|.$

Since an automorphism of $G$ should map a generator of $G$ to a generator of $G$, it's enough to know how many generators $G$ must have.

So $|\text{Aut}(G)|=|\mathbb{Z}_m^*| = \phi(m)$ where $\phi(m)$ is Euler's totient function.

(Note: you do not need that $m = p$, a prime; it suffices to know that $G$ is cyclic and finite.)

amWhy
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