In Pinter's "A Book of Abstract Algebra", the reader is asked to prove that the order of $a^m$ is equivalent to $\frac{{\rm lcm}(m,n)}{m}$ using the following two facts (which were proved in previous exercises):
- $(a^m)^{{\rm lcm}(m,n)*\frac{1}{m}}$ = e
- Given that $(a^m)^t=e$, ${\rm lcm}(m,n) \leq m*t$
Side note: the order for the element $a$ in this group is equal to $n$... ( ${\rm ord}(a)=n$).
My work so far has included the following steps:
By 1. , I know that ${\rm ord}(a^m) \leq \frac{{\rm lcm}(m,n)}{m}$.
By 2. , I know that $\frac{{\rm lcm}(m,n)}{m}\leq t$
This results in the following inequality:
${\rm ord}(a^m) \leq \frac{{\rm lcm}(m,n)}{m}\leq t$
And this is where I am stumped. I obviously need to somehow demonstrate that ${\rm ord}(a^m)$ cannot be less than $\frac{{\rm lcm}(m,n)}{m}$. However, if this is the case, I'm not quite certain why I need the proof from 2. to help establish this.
Any suggestions?
