Let \(n>1\) be an integer. Show that the automorphism group of the cyclic group of order \(n\) is isomorphic to the multiplicative group of units mod \(n\).
Solution
Let \(C_n=\langle g\rangle\) be cyclic of order \(n\) with generator \(g\).
1. For each \(k\in\ZZ\), the map \(\phi_k\colon C_n\to C_n\), \(\phi_k(g^a)=g^{ka}\), is a well-defined endomorphism, and every endomorphism of \(C_n\) equals \(\phi_k\) for some \(k\). Moreover \(\phi_k\) depends only on \(k \bmod n\).
If \(g^a=g^b\), then \(n\mid a-b\), so \(n\mid ka-kb\) and \(g^{ka}=g^{kb}\); thus \(\phi_k\) is well defined, and \(\phi_k(g^ag^b)=g^{k(a+b)}=\phi_k(g^a)\phi_k(g^b)\). An endomorphism \(\phi\) is determined by \(\phi(g)\), which equals \(g^k\) for some \(k\); then \(\phi=\phi_k\). Since \(\phi_k(g)=g^k\), the map \(\phi_k\) depends only on \(k\bmod n\).
2. The endomorphism \(\phi_k\) is an automorphism if and only if \(\gcd(k,n)=1\).
Since \(C_n\) is finite, \(\phi_k\) is bijective if and only if it is surjective, that is, if and only if \(g^k\) generates \(C_n\). The order of \(g^k\) is \(n/\gcd(n,k)\), which equals \(n\) exactly when \(\gcd(n,k)=1\).
3. The map \(\Psi\colon(\ZZ/n\ZZ)^\times\to\Aut(C_n)\), \(\Psi(k\bmod n)=\phi_k\), is a group isomorphism.
By steps 1 and 2, \(\Psi\) is well defined. For units \(k,\ell\), \((\phi_k\circ\phi_\ell)(g)=\phi_k(g^\ell)=g^{k\ell}=\phi_{k\ell}(g)\), so \(\phi_k\circ\phi_\ell=\phi_{k\ell}\) and \(\Psi\) is a homomorphism. If \(\phi_k=\operatorname{id}\), then \(g^{k-1}=e\) and \(k\equiv1\pmod n\), so \(\Psi\) is injective. By steps 1 and 2, every automorphism equals \(\phi_k\) with \(\gcd(k,n)=1\), so \(\Psi\) is surjective.
4. Q.E.D.
Step 3 gives \(\Aut(C_n)\cong(\ZZ/n\ZZ)^\times\).