I would like to see a proof, or at least to know when can I use that statement:
$$\text{ord}(mn,a)=\text{lcm}(\text{ord}(m,a),\text{ord}(n,a))$$
where
$\text{ord}(n,x)$ means the order of $x$ in $\mathbb{Z}_n $ (i.e the smallest $t$ such that $x^t = 1 \mod n$).
The identity does not always hold. Take for instance $m=n=2$ and $a=3$: the LHS becomes $2$ while the RHS is $1$. (There are a lot more counterexamples.)
However, if $\gcd(m,n)=1$, it will hold. Let $k=\text{ord}_{mn}(a)$, $x=\text{ord}_m(a)$ and $y=\text{ord}_n(a)$.
The relation $$a^{\text{lcm}(x,y)}\equiv1\pmod{mn}$$ follows since $a^{\text{lcm}(x,y)}\equiv1\pmod {m\text{ and }n}$ and $\text{lcm}(m,n)=mn$. Hence $k\mid\text{lcm}(x,y)$.
As $a^k\equiv1\pmod m$ and $a^k\equiv1\pmod n$ we have $x,y\mid k$. Therefore $\text{lcm}(x,y)\mid k$.
Combining $k\mid\text{lcm}(x,y)$ and $\text{lcm}(x,y)\mid k$ we obtain $k=\text{lcm}(x,y)$.
Note: The condition $\gcd(m,n)=1$ is sufficient, but not necessary to make the equality hold. If $a=1$ the identity becomes trivial, for example. Some more interesting examples can be constructed using base-$a$ Wieferich primes. For example, $m=n=1093$ and $a=2$ will make the inequality hold. Both sides then equal $364$. Also $m=n=3511$ and $a=2$ will work.
In fact, the precise condition is the following:
If $p$ is a common prime divisor of $m$ and $n$ with $p^s\|m$ and $p^t\|n$, then $$\text{ord}_{p^{s+t}}(a)=\text{ord}_{p^{\max(s,t)}}(a).$$
In the special case $s=1$, $t=1$ this can be shown to be equivalent to $$p^2\mid a^{p-1}-1,$$ which makes $p$ a base-$a$ Wieferich prime.
I think in general, similar relations can be obtained as in the case $s=t=1$. I'll search for it, though I think this might be difficult.
What is true is that $\ m\mid a^j-1,\ n\mid a^k-1\Rightarrow {{\rm lcm}(m,n)}\mid a^{{\rm lcm}(j,k)}-1.\,$ Simple counterexamples show that one cannot generally replace $\,{\rm lcm}(m,n)\,$ by $\,mn,\,$ e.g. let $\,j=1=k,\, m = a\!-\!1 = n > 1.$
Edit: I misread the question, so the following is pretty unrelated.
The general statements we can make here are the following.
Let $G$ be a finite abelian group and $a,b\in G$.
If anybody is interested, I can supply proofs for both statements.
