# Order of elements in a group

1. Jan 29, 2004

### Jupiter

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?

2. Jan 29, 2004

### phoenixthoth

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)...

3. Jan 30, 2004

### Hurkyl

Staff Emeritus
Here's a hint:

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

4. Jan 30, 2004

### Jupiter

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.

5. Jan 30, 2004

### Hurkyl

Staff Emeritus
Blarg, I always forget that step when I try to remember the proof to this problem!

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

6. Jan 30, 2004

### Jupiter

Hurkyl, your genuis quite disgusts me. In any event, I was able to complete the proof. Thanks so much for your help! You weren't by any chance born knowing this, were you?
In case you're interested, you have to instead compute (ab)^q=e^q. Once you get b^(qm)=e, you use the fact that b has order n and that n,m are relatively prime to show n|q. Similarly, you can show m|q. And so q is a common multiple of m,n, and so mn|q.