$U(p^m) \times U(q^n)$ is not cyclic

Question

Let $p$ and $q$ be odd primes and let $m$ and $n$ be positive integers. Explain why $U(p^m)\times U(q^n)$ is not cyclic.

Attempt-

$$ \begin{align*} U(p^m) \times U(q^n) &\cong \mathbb{Z}_{p^m-p^{m-1}} \times \mathbb{Z}_{q^n-q^{n-1}} \\ & \cong \mathbb{Z}_{p^{m-1}} \times \mathbb{Z}_{p-1} \times \mathbb{Z}_{q^{n-1}}\times \mathbb{Z}_{q-1} \end{align*} $$

Since $|\mathbb{Z}_{q-1}|$ and $|\mathbb{Z}_{p-1}|$ are not relatively prime , $U(p^m) \times U(q^n)$ is not cyclic.

Is my attempt correct?

Answer 1

You have the right idea but it can be simplified and made more precise:

Let $M=\phi(p^m)$ and $N=\phi(q^n)$. Then $U(p^m) \cong C_M$ and $U(q^n) \cong C_N$. Since $M$ and $N$ are both even, we have $\gcd(M,N) \ge 2$ and so $C_M \times C_N$ is not cyclic.

This is based on the following argument:

If $\gcd(M,N)>1$ then $C_M \times C_N$ is not cyclic because $g^L = 1$ for all $g \in C_M \times C_N$, where $L=\operatorname{lcm}(M,N)$. Since $L = \frac{MN}{\gcd(M,N)} < MN$, there is no element of order $MN$ in $C_M \times C_N$.