$\mathbf {The \ Problem \ is}:$ 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 $p.$


1 Answer 1


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.

  • 1
    $\begingroup$ I am glad that it cleared your doubt. Best wishes! $\endgroup$ Apr 11, 2020 at 19:16

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