$G$ finite cyclic group $\Rightarrow Aut(G) \cong \Bbb{Z}_n$

Question

Not sure if this question has been asked before, if so, let me know!

I need some help with the following

Problem. Let $G$ be a finite cyclic group with $|G|=n$. Prove that the group of all isomorphisms from $G$ to $G$ and the multiplicative group $\Bbb{Z}_n$ are isomorphic.

My attempt:

Suppose $G=\, <a>$ for some $a\in G$. Then the only generators of $G$ are $a^{k_1}, a^{k_2}, \dots, a^{k_{\phi(n)}}$, where $1\le k_i\le n$, $1\le i\le \phi(n)$, $(k_i,n)=1$ and $\phi$ is Euler's totient funtion. What I've noted is that if $f\in Aut(G)$ then $f(G)=G=<f(a)>$. Hence $f(a)=a^{k_i}$ for some $1\le i\le \phi(n)$.

What I'm trying to do above is to find a way to characterize the elements of $Aut(G)$... And trying to convince myself that $|Aut(G)|=|\Bbb{Z_n}|$.

I don't know where to go from here. Also, I don't see how $|Aut(G)|=|\Bbb{Z_n}|=\phi(n)$.

Help and hints are highly appreciated.

Answer 1

As $G$ is cyclic, a homomorphism $f$ is completely specified once $f(a)$ is specified, $a$ being a generator. It has to be $a^k$ for some $k$, with $k<n$ This homomorphism will be an automorphism iff $\gcd(k,n)=1$. Hence the order of automorphism group is $\phi(n)$.

Answer 2

Let $G=\langle a \rangle$ be a cyclic group of order $n$ and $\phi $ be an automorphism of $G$.

$\phi(a)=a^k$ for some $k\in \mathbb{N}$.
Also, $\phi(a^l)=a$ for some $l\in \mathbb{N}$.
Combining the two results, we can get $a^{kl}=a$.
So $n\mid kl-1$ which means $kl+cn=1$ for some $c\in \mathbb{Z}$.

Here we need to use some result from number theory, which is Bezout's Theorem.
Since $k(l)+n(c)=1$, we conclude that $(k,n)=1$. So the possible values for $k$ is $\varphi(n)$ where $\varphi(n)$ denotes the Euler's totient function of $n$.

Extended idea of solution:
From previous part, we can write $Aut(G)=\{\phi_{k_1},\phi_{k_2},\dots,\phi_{k_{\varphi(n)}}\}$ where $\phi_{k_i}$ is an automorphism defined by sending each element $x\in G$ to $x^{k_i}$ where $(k_i,n)=1$ for each $i$.

Define another map $\alpha:Aut(G) \rightarrow \mathbb{Z}_n^\times$ by $$\alpha(\phi_{k_i})=k_i$$ So it can be verified that $\alpha$ is an isomorphism.

(Take note that here I use $\varphi$ for Euler totient function, not $\phi$, sorry for the confusion arised)

Answer 3

Let's assume $G=\mathbb{Z}/n\mathbb{Z}$, with $n\ge0$.

Consider $\pi\colon\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$, the canonical projection. If $f\colon\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ is an endomorphism, then $\tilde{f}=f\circ\pi$ is determined by $\tilde{f}(1)$:

if $a=\tilde{f}(1)$, then $\tilde{f}(z)=za$, for every $z\in\mathbb{Z}$.

Note that $n\mathbb{Z}\subseteq\ker\tilde\varphi$, so, in particular, $na=0$.

Conversely, given $a\in\mathbb{Z}/n\mathbb{Z}$, the map $g\colon\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $g(z)=za$ is a homomorphism and $\ker g\supseteq n\mathbb{Z}$, so $g$ induces uniquely a homomorphism $g_0\colon \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ such that $g_0\circ\pi=g$.

How does $g_0$ act? If $a=g_0(1+n\mathbb{Z})=g(1)$, then, for $x\in\mathbb{Z}/n\mathbb{Z}$, $g_0(x)=ax$ (the product in $\mathbb{Z}/n\mathbb{Z}$), as it is easy to verify.

This means that every $a\in\mathbb{Z}/n\mathbb{Z}$ uniquely defines a homomorphism $\mu_a\colon\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ such that $\mu_a(x)=ax$.

It's clear that, for $a,b\in\mathbb{Z}/n\mathbb{Z}$, we have $\mu_a\circ\mu_b=\mu_{ab}$. Thus the endomorphisms of $\mathbb{Z}/n\mathbb{Z}$ are in bijection with the elements of $\mathbb{Z}/n\mathbb{Z}$ and composition corresponds to multiplication.

In particular the automorphisms of $\mathbb{Z}/n\mathbb{Z}$ are of the form $\mu_a$ where $a$ is invertible in $\mathbb{Z}/n\mathbb{Z}$ (under multiplication).

For $n=0$, this means that the automorphisms of $\mathbb{Z}$ correspond to the invertible elements of $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so $\mu_{1}$ and $\mu_{-1}$.

For $n>0$, the invertible elements of $\mathbb{Z}/n\mathbb{Z}$ under multiplication are those which are of the form $x+n\mathbb{Z}$ with $\gcd(x,n)=1$. Thus the automorphism of $\mathbb{Z}/n\mathbb{Z}$ are in bijection with $U(\mathbb{Z}/n\mathbb{Z})$ (the group of units) and their number is $\varphi(n)$ (Euler's totient function).