Let $p,q > 2$ be distinct primes. Show that $\mathbb{Z}_{pq}^*$ is not cyclic. I recognize this question has been asked many times before (here, here, and here) and in other forms (that $pq$ does not have a primitive root for example).
I am also self-studying Aluffi's Algebra and am wondering specifically about the last solution above I have linked.
We know that $\mathbb{Z}_{pq}^*$ has order $pq$ less the multiples of $p$ or $q$ between $1$ and $pq$. These multiples of $p$ are $p,2p,...,(q-1)p$, of $q$ are $q,2q,...,(p-1)q$, and of course their least common multiple $pq$. So the order is $pq - (p-1) - (q-1) - 1 = (p-1)(q-1)$.
We aim to show that there is no element of this order so that the group cannot be cyclic. Put $n = (p-1)(q-1) /2$. Both factors in the numerator are even so we can do this and still have both factors dividing $n$.
The last solution linked above uses that for $m\in \mathbb{Z}_{pq}^{*}$ we have $m$ in both $\mathbb{Z}_{p}^{*}$ and $\mathbb{Z}_{q}^{*}$, with 
$$ m^{n} \equiv m^{p-1} \equiv 1 \mod p  \\
m^{n} \equiv m^{q-1} \equiv 1 \mod q.$$
From which we get $p,q | m^{n} - 1$ so $pq | (m^{n}-1)$ since $p,q$ are coprime. But how do we know that
$$ m^{p-1} \equiv 1 \mod p  \\
m^{q-1} \equiv 1 \mod q$$
without something like Lagrange's theorem? Aluffi doesn't develop a lot for us to work with...
 A: I believe I have solved it. 
Let $m\in\mathbb{Z}_{pq}^{*}$. We have that $m\in\mathbb{Z}_{p}^{*}$ and $m\in\mathbb{Z}_{q}^{*}$. Let $k$ and $l$ be the orders of $m$ in these latter groups, resp. We see that $k \leq p-1$ and $l \leq q-1$ because the groups only have that many elements. If equality holds in both cases, then the result follows from the considerations in the question above. So allow one of the inequalities to be strict. 
We see that 
$$m^{kl} \equiv (m^{k})^{l} \equiv 1 \mod p \\
m^{kl} \equiv (m^{l})^{k} \equiv 1 \mod q$$
so we must have $m^{kl} \equiv 1 \mod pq$. Thus the order of $m \in \mathbb{Z}_{pq}^{*}$ is at most $kl$. Since $kl < (p-1)(q-1)$, we are done.
A: You know that $\mathbb Z_p^*$ is of order $p-1$ because $p$ is prime. But group theory tells us that the order of any element in a group divides the order of the group itself.
So if $m\in \mathbb Z$ is not a multiple of $p$ (which is the case if you assume that the classe of $m$ in $\mathbb Z_{pq}$ is invertible), you have $$ m^{p-1} \equiv 1 \mod p$$
