# Order of elements in a group

If G is an abelian group, a in G has order m, b in G has order n, and gcd(m,n)=1, show that ab has order mn.

I am able to show that (ab)^(mn)=e actually occurs. I am having great difficulty however showing that mn is the smallest such integer. I tried to assume that there were a smaller integer, but I could not derive any contradiction. I tried to use the fact that gcd(m,n)=1 as best I could, but I can't make it work for me.

Anyone have any ideas on where to go with this proof?

first of all, suspect that you have a minor detail of treating the cases when at least one of a and b equals e, having order 1.

i tried this for a while and didn't make heads or tails of it.

then i did the always helpful example. i supposed m=o(a)=4 and n=o(b)=15. note that gcd(4,15)=1.

suppose o(ab)=k<mn=60.

then (ab)^k=e.

as an example, i supposed k=59.

note that a^59=a^3 because o(a)=4 and 59 = 3 mod 4.

also note that b^59=b^14 because o(b)=15 and 59 = 14 mod 15.

by assuming k=59, ie o(ab)=59, (ab)^59=e.

e=(ab)^59=a^59 b^59 (as G is abelian)
=a^3 b^14.

now "multiply" both sides by ab (using abelian again) to get this:
ab=a^4 b^15=e. hence o(ab)=1<k=59 but k was supposed to be the smallest "power" for which (ab)^k=e

now the harder part is proving that it is true in general. one thing i didn't seem to explicitly make use of is the relative primality of o(a) and o(b)...

Hurkyl
Staff Emeritus
Gold Member
Here's a hint:

gcd(m, n) = 1 if and only if lcm(m, n) = mn.

Hurkyl, your hint would be very useful if I could somehow show that the order of ab must be a common multiple of m and n.

.... which I cannot do. I suppose that ab has order k, and then I try to show n|k and m|k. I then suppose that I do not have n|k and m|k. I apply the division algorithm and get a^r1b^r2=e. And this does me no good unless I can show ri=0.

Hurkyl
Staff Emeritus
The trick is, IIRC, that $a^p b^q = e$ implies $b^{qm} = e$, and do something from there.