- #1

erogard

- 62

- 0

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

Now we define [tex] G_{m} = [/tex] { [tex] g \in G | [/tex] order(g) divides m }.

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

So I let [tex] k_{1} [/tex] and [tex] k_{2} [/tex] 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.