If $m = pq$ is the product of two distinct odd primes $p$ and $q$, prove that $ord(x)|φ(m)/2$ for all $x ∈ (Z/mZ) ^*$ If $m = pq$ is the product of two distinct odd primes $p$ and $q$, prove that $ord(x)|φ(m)/2$ for all $x ∈ (Z/mZ)^*$. Conclude that $(Z/mZ)^*$ is not cyclic. 
The order of $(Z/pZ)^*$ and $(Z/qZ)^*$ is $p-1$ and $q-1$ respectively. So order of $(Z/mZ)^*$ should be $(p-1)(q-1)=φ(m)$. How do we get to $ord(x)|φ(m)/2$?
And is the group not cyclic because $p$ and $q$ are primes so it doesn't have one generator?
 A: Because of Fermat's theorem, we have 
$x^{p-1} \equiv 1 \bmod p$ and so $(x^{p-1})^{\frac{q-1}{2}} \equiv 1 \bmod p$.
Analogously, $(x^{q-1})^{\frac{p-1}{2}} \equiv 1 \bmod q$.
Therefore, $x^{\frac{(p-1)(q-1)}{2}}  \equiv 1 \bmod pq$.
This implies that $ord(x) \le \frac{(p-1)(q-1)}{2}=\frac{\varphi(m)}{2} < \varphi(m)$.
Thus, no element of $(\mathbb Z / m \mathbb Z)^\times$ can have order $\varphi(m)$ and the group cannot be cyclic.
A: The answer given by @Ihf can be be more general and more precise at the same time. If $m=ab$ with co-prime $a, b$ , by the Chinese Remainder theorem, $\mathbf Z / m\mathbf Z \cong \mathbf Z / a\mathbf Z \times \mathbf Z / b\mathbf Z$ as rings, so $(\mathbf Z / m\mathbf Z)^* \cong (\mathbf Z / a\mathbf Z)^* \times (\mathbf Z / b\mathbf Z)^*$ as groups. In such a direct product of groups, if $x=(x_1, x_2)$, obviously $ord x = l.c.m.(ord x_1, ord x_2) =  (ord x_1. ord x_2)/g.c.d. (ord x_1, ord x_2)$. This shows also that $(\mathbf Z / m\mathbf Z)^*$ is cyclic iff $(\mathbf Z / a\mathbf Z)^*$ and $(\mathbf Z / b\mathbf Z)^*$ are cyclic and moreover $\phi (a)$ and $\phi (b)$ are co-prime. In your special case, $2$ is a c.d. of $p-1$ and $q-1$, so $ord x$ divides $(p-1)(q-1)/2$, and $(\mathbf Z / m\mathbf Z)^*$ is never cyclic.
