can composite groups have primitive roots (be cyclic)?

Question

Imagine Z/nZ with n not prime (n=pq). Can the multiplicative group be cyclic?

I read the paper over here: http://alpha.math.uga.edu/~pete/4400primitiveroots.pdf

At one point he dismiss the case where the group can be isomorphic to the product of several groups:

Thus it is enough to figure out the group structure when $N=p^a$ is a prime power.

Is it because a product of group can't be cyclic?

Thanks!

Answer 1

Yes, this can happen for $p=2$ and $q>2$ another prime. In fact, there is a primitive root modulo $n$ if and only if $n$ is $1$ or $2$ or $4$ or $p^\alpha$ or $2p^\alpha$, where $p$ is prime, $p\ne2$, and $\alpha\ge1$, see Corollary $2$ of P. Clark's text. For $n=pq$ with $p>2,q>2$ both prime there is hence no primitive root, so that $(\mathbb{Z}/pq)^*$ is not cyclic in this case.

Answer 2

OK ended up finding why:

if $(Z_n)^\ast \sim (Z_p)^\ast \times (Z_q)^\ast$ (the simplest case with $p$ and $q$ primes)

then the order of $(Z_p)^\ast$ is $p-1 = 2 \times \cdots$,

and the order of $(Z_q)^\ast$ is $q-1 = 2 \times \cdots$,

this means the gcd of their order will not be equal to 1, this means the product group won't be cyclic. Explanation about that part here