Question:
Let p and q be odd primes and m and n be positive integers. Explain why $U\left ( p^{m} \right ) \bigoplus U\left ( q^{n} \right )$ is not cyclic.
Note that $U\left ( n \right )$-group is the group of integers less than and relatively prime to n.
Recall that $U\left ( p^{n} \right )\cong \mathbb{Z}_{p^{n}-p^{n-1}}$.
Then, $U\left ( p^{m} \right ) \bigoplus U\left ( p^{n} \right ) \cong \mathbb{Z}_{p^{m}-p^{m-1}} \bigoplus \mathbb{Z}_{q^{n}-q^{n-1}}$.
$\mathbb{Z}_{n}$ can be generated by 1 so this is a cyclic group. Following this, both $\mathbb{Z}_{p^{m}-p^{m-1}}$ and $\mathbb{Z}_{q^{n}-q^{n-1}}$ are finite cyclic group.
Clearly, the order of $\mathbb{Z}_{p^{m}-p^{m-1}} is p^{2m-1}$ and the order of $\mathbb{Z}_{q^{n}-q^{n-1}}$ is $q^{2n-1}$
It suffices for me to show that the orders are relatively prime.
Any hints are appreciated.
Thanks in advance.
We have $\left|U\left ( p^{m} \right )\right| = p^{m-1} (p-1)$ and $\left|U\left ( q^{n} \right )\right| = q^{n-1} (q-1)$ by totient.
Since $p$ and $q$ are odd primes, their orders are both even. Let the orders be $2h$ and $2k$ respectively, where $h$ and $k$ are positive integers.
For the sake of a contradiction, assume that $U\left ( p^{m} \right ) \bigoplus U\left ( q^{n} \right )$ is generated by $(a,b)$.
The order of $a$ must be $2h$, and the order of $b$ must be $2k$, meaning that the order of $(a,b)$ is $2\gcd(h,k)$.
Notice that $(a,b^2)$ is not a power of $(a,b)$. For if $(a,b)^m = (a,b^2)$, then $m=2hr$ and $m=2ns+1$, so $m$ is both even and odd, a contradiction.
Since $(a,b^2)$ is not a power of $(a,b)$, $(a,b)$ is not a generator of $U\left ( p^{m} \right ) \bigoplus U\left ( q^{n} \right )$, which is hence not cyclic.