A short conjecture on primitive roots I was studying elementary number theory and I made myself the following question: If $m = p.q$, where $p,q$ are distinct primes, which is the maximum value of $k$ such that there is an element over $Z_m$ with order k?
What I did to solve the problem was:
I was trying to make a proof using the following arguments, some of then I can't prove:
First of all, if $r$ is an element $Z_m$ then $|r|$ has to be multiple of $p-1, q-1$.
(I can't prove this)
Second, if $a$ is coprime with $m$ then choosing $n = \operatorname{lcm}(|a|_p,|a|_q)$, where $|a|_q$ is the order of $a$ over $Z_q$ then
$a^n \equiv 1 \pmod m$. 
This has a easy proof.
Third, If $a$ is a solution of $x\equiv r \pmod p$ and $x\equiv 1 \pmod q$ then $|a| = p-1.$
What I did was using the Chinese remainder theorem.
And the last statement:
$\operatorname{max}\{k \in Z : \text{there is} ~c~ \text{with order}~ k\} = \operatorname{lcm}(p-1,q-1).$
Is quite obvious this statement but I am not so sure if I can prove.
Once all of this is true, can I assure that:
If $m = \prod_{j=1}^rp_j^{\alpha_j}$, then the maximum value of $k$ such that there is $a \in Z_m$ such that $|a| = k$ is given by $\operatorname{lcm}(\phi(p_1^{\alpha_1}),\ldots,\phi_r(p_r^{\alpha_r})).$

You're right, except when $m$ is even, in which case you need to use $\phi(2^e)/2$ in that lcm.
This is the role of the Carmichael function $\lambda(n)$.
We always have that $\lambda(n)$ divides $\phi(n)$. Typically $\lambda(n)$ is much smaller than $\phi(n)$.
