# Homework Help: Unique Elements of Relatively Prime Order

1. Oct 12, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Let G be a group and let g be an element G such that |g| = mn, where gcd(m, n) = 1. Show that there are unique elements a, b in G such that g = ab = ba and |a| = m and |b| = n.

The attempt at a solution
Since gcd(m, n) = 1, there are integers s and t such that ms + nt = 1. Hence g = gms+nt = gmsgnt. Let a = gnt and b = gms. Then g = ab = ba. We also have that |a| ≤ m and |b| ≤ n.

I'm having trouble showing that |a| = m and |b| = n. Also, the fact that s and t are not unique proves that a and b are not unique right? This is as my thought process has taken me. Any tips?

2. Oct 12, 2008

### tiny-tim

Hi e(ho0n3!

s and t are unique (to prove it, assume they're not).

And |a| must be a factor of mn, so … ?

3. Oct 12, 2008

### e(ho0n3

I don't think so: Take m = 7 and n = 11. Then gcd(7, 11) = 1 and 7(-3) + 11(2) = 7(8) + 11(-5) = 1.

Why must it be a factor of mn?

4. Oct 12, 2008

### morphism

Notice that |g^m| = |g|/gcd(|g|, m) = n. So |g^(ms)|=<fill blank>. Similarly, |g^(nt)|=<fill blank>.

So now you've found the appropriate a and b. To show uniqueness, suppose that g=a'b'=b'a' and |a'|=m and |b'|=n, then show that a'=a and b'=b. The fact that |a'|=|a|=m and |b'|=|b|=n is important here.

5. Oct 12, 2008

### e(ho0n3

We have that |b| = mn / gcd(mn, ms) = n / gcd(n, s) and |a| = mn / gcd(mn, nt) = m / gcd(m, t). Are you suggesting that gcd(n, s) = gcd(m, t) = 1 because ms + nt = 1?

Concerning uniqueness, I've managed to get bm = (b')m from (ab)m = (a'b')m and similarly an =(a')n. How do I continue.

6. Oct 12, 2008

### CoCoA

If gcd(n,s)>1, what could you say about ms + nt?

7. Oct 12, 2008

### e(ho0n3

I could say that gcd(n,s) divides ms + nt since it divides ms and nt, but this means it would divide 1, which is impossible. I get it know.

Somehow I need to get rid of the n in an = (a')n. I'm thinking of the following: Consider the cyclic group <a>. Since <a> has order m and n is relatively prime to m, we must have <an> = <a>. We also get that <(a')n> = <a'>, hence <a> = <a'> and a = a'?

8. Oct 12, 2008

### morphism

You already used the trick required here: b = bnt bms.

9. Oct 12, 2008

### e(ho0n3

Ah, right! That hadn't occured to me. Thanks a lot.