Automorphism of a group of integer is isomorphic to U(n)

Question

Theorem:

For every $n>1$,

$\textrm{Aut}(\mathbb{Z}_{n}) \cong \textrm{U}_n.$

The proof begins with the consideration:

If $\textrm{gcd}(n, k) = 1$ then $\phi_{k}:\mathbb{Z}_{n}\rightarrow \mathbb{Z}_{n}$

$m \mapsto mk \pmod n$ is an automorphism of $\textrm{U}_n$.

I'm a bit lost as to why the map is defined in such a way. That is, from $m$ to $m \cdot k \pmod n$.

Could someone shed some light?

Answer 1

An automorphism of $\mathbb{Z}_n$ is determined by its image on $1$. So define $\phi_k : \mathbb{Z}_n \to \mathbb{Z}_n$ by $\phi_k(1) = k$. This map is an automorphism if and only if $\gcd(n,k) = 1$. To see this, note that $\phi_k$ is an automorphism if and only if $\left<k\right> = \mathbb{Z}_n$, i.e. $k$ has order $n$ in $\mathbb{Z}_n$. Since the order of an element $a$ in $\mathbb{Z}_n$ is given by $n/\gcd(a,n)$, this happens if and only if $\gcd(n,k) = 1$.

Answer 2

Every cyclic group of cardinality $n$ is isomorphic to $\mathbb Z_{n}$. Then if we look at the generators of $\mathbb Z_{n}$, we see cyclic groups generated by them must be isomorphic to $\mathbb Z_{n}$.

For example, let's examine $\mathbb Z_{8}$. In this case we have $4$ generators: $1, 3, 5, 7.$ They are all coprime to $8$. Indeed, a residue class $a$ generates $\mathbb Z_{n}$ iff $(a, n) = 1$. Since this is the only way $\textrm{lcm}$ of $a$ and $n$ can be equal to $an$, the case that $n$ distinct elements are generated. So $U_{n}$ is the set of all generators.

Since those cyclic groups are just permutations of $\mathbb Z_{n}$, we can consider the permutations to be automorphisms of $\mathbb Z_{n}$.

The identity automorphism for $\mathbb Z_{8}$ (repeatedly add residue class $1$ modulo $8$ until it repeats):

$1 := (0, 1, 2, 3, 4, 5, 6, 7)$

Some other example:

(repeatedly add residue class $3$ modulo $8$ until it repeats):

$3 := (0, 3, 6, 1, 4, 7, 2, 5)$

$5 := (0, 5, 2, 7, 4, 1, 6, 3)$

And so on. So we can say that the function $\varphi = ax \pmod n$ is an automorphism of $\mathbb Z_n$ for $\forall a \in U_n$. And set of all such functions forms the automorphisms group of $\mathbb Z_n$.