Can there be involutions in the unit group of $\mathbb{Z}_n$?

Question

Let $x\neq 1$ be an element of the multiplicative group $\mathbb{Z}_n^*$. Then my question is, is it possible for $x^2$ to equal $1$?

For what $n$ is it possible or impossible?

Answer 1

Firstly, and most importantly, your title is incorrect. Idempotents are elements $x\in G$ such that $x^2=x$, and it is an exercise in elementary group theory that the only idempotent in a group is the identity. As has been mentioned in the comments, it looks like you mean to say involutions instead of idempotents, so I answer on this assumption.

In general this is very possible, consider, for example, $4$ modulo $15$ (or indeed $(-1)^2=1$ always as is noted above). The important thing is that involutions are elements of order $2$. The general structure of $\mathbb{Z}/n\mathbb{Z}^\times$ is well known, (see e.g. the wikipedia page or most good introduction to groups/number theory books). I give it below:

Firstly, for an odd prime $p$ and positive integer $k$ we have $$\mathbb{Z}/p^k\mathbb{Z}^\times\cong \mathbb{Z}/(p^k-p^{k-1})\mathbb{Z}.$$

Secondly, at powers of $2$ we have $$\mathbb{Z}/2^k\mathbb{Z}^\times\cong \begin{cases} \{1\}&k=1\\ \mathbb{Z}/2&k=2\\ \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2^{k-2}\mathbb{Z} &k\geq 3 \end{cases}$$

The general case then follows from the Chinese remainder theorem, which says that if $n=\prod_{i=1}^rp_i^{k_i}$ is the unique prime decomposition of $n$ then $$\mathbb{Z}/n\mathbb{Z}^\times=\prod_{i=1}^r\mathbb{Z}/p_i^{k_i}\mathbb{Z}^\times$$

From the structure of the unit group as a product of cyclic groups in this way, it is easy to see how often (and even how many) involutions appear.