Showing that a subset is a group

[erogard]

Hi,

Consider a finite, additive and abelian group $$G$$ of order $$mn$$, where $$gcd(m,n)=1$$. Basically m an n are relatively prime.

Now we define $$G_{m} =$$ { $$g \in G |$$ order(g) divides m }.

I'm trying to show that $$G_{m}$$ is a subgroup of G. All I really need to show is that it is closed under addition. That is, $$g_{1}, g_{2} \in G \Rightarrow g_{1} + g_{2} \in G$$.

So I let $$k_{1}$$ and $$k_{2}$$ be the order of g1 and g2, respectively, and try to express the order of (g1 + g2) in terms of k1 & k2, and, moreover, to show that order (g1+g2) divides m.

I know that k1k2 does the job, that is, k1k2(g1 + g2) = e, but k1k2 may not divide m since we might end up with k1k2 > m.

I've tried a lot of different things, so far unsuccessfully, but any ideas or suggestions would be much appreciated.

 

[Dick]

How about lcm(k1,k2) (least common multiple). That works doesn't it? It's certainly less than m.

 

[erogard]

How about lcm(k1,k2) (least common multiple). That works doesn't it? It's certainly less than m.

Thanks, that certainly does it - I'm just trying to convince myself that it is less than m.

 

[Dick]

Thanks, that certainly does it - I'm just trying to convince myself that it is less than m.

Then yourself is hard to convince. m IS a common multiple of k1 and k2.

 

[erogard]

Then yourself is hard to convince. m IS a common multiple of k1 and k2.

haha, right, I think I've been a bit slow thinking last night. Thanks for your help.