Show that $(\mathbb{Z}/n)^\times$ cannot be cyclic unless $n$ is either a prime power or twice a prime power.

Q: Show that $(\mathbb{Z}/n)^\times$ cannot be cyclic (i.e., there is no primitive root modulo $n$) unless $n$ is either a prime power or twice a prime power.

I'm reading a solution. The first step is that for any $n$ it can be factorized into three relatively prime terms: $n = 2^r \cdot p^s \cdot m$ where $p$ is prime.

Since these terms are relatively prime, we can say that:

$(\mathbb{Z}/n)^\times \cong (\mathbb{Z}/2^r)^\times \times (\mathbb{Z}/p^s)^\times \times (\mathbb{Z}/m)^\times$

The next step claims that for $(\mathbb{Z}/n)^\times$ to be cyclic, at most one of the three factored groups can be an even cyclic group. Why is this statement true? I know what cyclic groups are, does "even cyclic" just mean a cyclic group of even order?

First, a general fact about finite groups:

If $G = A \times B$ and $gcd(|A|,|B|)>1$, then $G$ cannot be cyclic.

Indeed, let $m=lcm((|A|,|B|)$. Then $x^m =1$ for all $x\in G$. But $m<|A|\cdot|B|=|G|$ because $gcd(|A|,|B|)>1$. Therefore, no element in $G$ can have order $|G|$ and $G$ cannot be cyclic.

$(\mathbb{Z}/p^s)^\times$ has order $p^{s-1}(p-1)$ and so is even if $p$ is odd.

As a consequence, if $n$ has two or more odd prime factors, then $(\mathbb{Z}/n)^\times$ is not cyclic because it will have two direct factors of even order.

The same conclusion holds if $r>1$ in $n = 2^r \cdot p^s \cdot m$.