\(\newcommand{\ZZ}{\Bbb Z} \newcommand{\ZZn}[1]{\ZZ / {#1} \ZZ}\)

One of the most familiar rings is the ring of integers modulo \(n\), often denoted \(\ZZn{n}\). Like all rings, it has an additive structure and a multiplicative one. The additive structure is straightforward: \(\ZZn{n}\) is cyclic, generated by \(1\). In fact, every integer \(a\) coprime to \(n\) is a generator for this group, giving a total of \(\phi(n)\) generators. The multiplicative structure, on the other hand, is far less apparent.

Not all elements of \(\ZZn{n}\) can participate in the multiplicative group, because not all of them have inverses. For example, 4 has no inverse in \(\ZZn{6}\); there’s no integer \(a\) such that \(4a \equiv 1 \pmod 6\). Elements that do have inverses are called units, and we’ll denote the group of units in \(\ZZn{n}\) as \(U_n\).

Since an element \(a \in \ZZn{n}\) is a unit iff \(a\) and \(n\) are coprime, there are \(\phi(n)\) units, where \(\phi\) is the totient function. But the size of the group alone doesn’t nail down the group structure.

For example:

• \(U_5 = \{ 1, 2, 3, 4 \}\):
  • generated by \(2\): \(2^0 = 1\), \(2^1 = 2\), \(2^2 = 4\), \(2^3 = 8 = 3\)
  • also generated by \(3\): \(3^0 = 1\), \(3^1 = 3\), \(3^2 = 4\), \(3^3 = 2\)
  • this group is isomorphic to \(\ZZn{4}\)
• \(U_8 = \{ 1, 3, 5, 7 \}\)
  • every element squares to \(1\)
  • this group is isomorphic to \(\ZZn{2} \times \ZZn{2}\)

Is there a way to find the structure of \(U_n\)?