# A problem regarding the product of all the elements of $U_n$ for some selected $n$

$$\mathbf {The \ Problem \ is}:$$ Find the product of all elements of the multiplicative group $$U_n$$ where $$n=p^2q$$ and $$p^2$$ for distinct primes $$p$$ and $$q ?$$

$$\mathbf {My \ approach} :$$ Actually, product of all the elements of a finite commutative group is $$1$$ unless there is any element of order $$2.$$ But, for the case $$n=2^2.3=12$$, note that each element of $$U_{12}$$ is of order $$2$$, and the product becomes $$1$$ , and we know $$U_{p^2}$$ is cyclic, but taking $$p=2,3$$ etc., it turns out to be $$-1.$$

If the total number of elements of order $$2$$ can be figured out, then only their product can be dealt differently. But, I can't approach further .

A small hint is warmly appreciated.

$$\mathbf {Edit}:$$ I think we can use the fact that $$U_n$$ is a cyclic group for $$n=1,2,4,p^k,2p^k$$ for any prime $$p.$$

Notation: If $$R$$ is ring with 1, let $$U(R)$$ denote the multiplicative group of all units in $$R$$. So $$U_n=U(\mathbb{Z}/n\mathbb{Z})$$ for any positive integer $$n$$.
Let $$m,n\in \mathbb{N}$$ such that $$(m,n)=1$$. Then the Chinese remainder theorem allows us to conclude that $$\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/mn\mathbb{Z}$$ as rings. Therefore the units correspond in a one-to-one way via this isomorphism, ie., if the isomorphism is denoted by $$\phi$$, $$\phi|_{U(\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z})}$$ is a group isomorphism onto $$U(\mathbb{Z}/mn\mathbb{Z})$$.
Now, $$U(\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z})=U(\mathbb{Z}/m\mathbb{Z})\times U(\mathbb{Z}/n\mathbb{Z})$$.
In the situation of the question asked here, it is now possible to precisely locate what are all the order two elements in the group $$U_{p^2q}$$. I hope that this helps.