1. ## Abelian Group Question

G is an abelian group.

a,b are in G
order of a = m
order of b = n

what is this order of ab ?

Struggling with this. I suspect it is lcm(m,n) but unable to prove it.

2. Originally Posted by aman_cc
G is an abelian group.

a,b are in G
order of a = m
order of b = n

what is this order of ab ?

Struggling with this. I suspect it is lcm(m,n) but unable to prove it.

it's not necessarily lcm(m,n). the only thing we can say is that $o(ab)$ divides lcm(m,n). the point is that $o(ab)$ doesn't only depend on $o(a)$ and $o(b)$. it also depends on the relationship between $a$

and $b$. for example if $b=a^{-1},$ then $o(a)=o(b)=m$ but $o(ab)=1.$

3. Originally Posted by NonCommAlg
it's not necessarily lcm(m,n). the only thing we can say is that $o(ab)$ divides lcm(m,n). the point is that $o(ab)$ doesn't only depend on $o(a)$ and $o(b)$. it also depends on the relationship between $a$

and $b$. for example if $b=a^{-1},$ then $o(a)=o(b)=m$ but $o(ab)=1.$
Thanks! I get it.

I had worked out the following some time back
$
\frac{mn}{\gcd(m,n)^2}|o(ab)$

Does it make sense?

4. Originally Posted by aman_cc
Thanks! I get it.

I had worked out the following some time back
$
\frac{mn}{\gcd(m,n)^2}|o(ab)$

Does it make sense?
that's correct. so we have $\frac{\text{lcm}(m,n)}{\gcd(m,n)} \mid o(ab) \mid \text{lcm}(m,n).$ this is not bad but to find $o(ab),$ as i mentioned, we need to know how $a,b$ are related.