Multiplicative Order of a number as the modulo changes So it is a well-known fact that
$${\rm lcm}({\rm ord}_n(a), {\rm ord}_m(a)) ={\rm ord}_{mn}(a)$$
when $\gcd(m, n) = 1$. After lots of experiments, and I suspect that when $\gcd(m, n) \neq 1$,
$${\rm ord}_{mn}(a) = {\rm lcm}({\rm ord}_n(a), {\rm ord}_m(a)) \cdot \gcd(m, n).$$
I attempted to prove it but wasn't able to and I want to know if this is true in general.
edit: As was pointed out, the argument obviously does not work when $a = 1$ and most of my experiment assumed $a = 2$ since it is the easiest. But I still suspect that my conjecture holds when $a \neq 1$.

When $\gcd(m,n) \ne 1$, no such formula can hold, because $\mathrm{ord}_m(a)$ and $\mathrm{ord}_n(a)$ do not necessarily determine $\mathrm{ord}_{mn}(a)$.
For example, take $m=15$ and $n=21$. Then:

*

*$\mathrm{ord}_{15}(8) = 4$, $\mathrm{ord}_{21}(8) = 2$, and $\mathrm{ord}_{15\cdot 21}(8) = 4$.

*$\mathrm{ord}_{15}(13) = 4$, $\mathrm{ord}_{21}(13) = 2$, but $\mathrm{ord}_{15 \cdot 21}(13) = 12$.

What we know is that if $a^k \equiv 1 \pmod{mn}$, then $a^k \equiv 1 \pmod m$ and $a^k \equiv 1 \pmod n$; therefore $\mathrm{ord}_m(a) \mid \mathrm{ord}_{mn}(a)$ and $\mathrm{ord}_n(a) \mid \mathrm{ord}_{mn}(a)$; hence $$\mathrm{lcm}(\mathrm{ord}_n(a), \mathrm{ord}_m(a)) \mid \mathrm{ord}_{mn}(a).$$ But the quantity $\frac{\mathrm{ord}_{mn}(a)}{\mathrm{lcm}(\mathrm{ord}_n(a), \mathrm{ord}_m(a))}$ can vary.
