Order of elements in a group

  • Thread starter Jupiter
  • Start date
46
0

Main Question or Discussion Point

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?
 

Answers and Replies

1,569
1
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
Science Advisor
Gold Member
14,843
17
Here's a hint:

gcd(m, n) = 1 if and only if lcm(m, n) = mn.
 
46
0
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.

Please Hurkyl a further hint.
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
Blarg, I always forget that step when I try to remember the proof to this problem!


The trick is, IIRC, that [itex]a^p b^q = e[/itex] implies [itex]b^{qm} = e[/itex], and do something from there.
 
46
0
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.
 

Related Threads for: Order of elements in a group

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
2
Views
7K
Replies
8
Views
6K
Replies
4
Views
2K
Replies
10
Views
9K
Replies
7
Views
4K
Replies
10
Views
7K
Top