# Groups and subgroupd

1. Nov 29, 2005

### TimNguyen

Let (G,*) be a group and (S,*) a subgroup of G. Prove that if for an element a in G, there exists m,n in Z, which are relatively prime, such that a^m and a^n is in S, then a is in S.

At the moment, I think the problem is trivial but something just tells me it is not.

2. Nov 30, 2005

### Muzza

What do you know about relatively prime integers? There's a well-known characterization of them which could be used here.

3. Nov 30, 2005

### TimNguyen

So m and n have no common factors?
Thus if a^m is in S and a^n is in S, a^m share no common divisors with a^n and vice versa so a must be in S?

4. Nov 30, 2005

### matt grime

It doesn't make sense to talk about divisors in a group like that. What does it mean to talk about common divisors of two rotations of a polygon?

It isn't that m and n have no common factors: they do, all numbers have common factors, and one called the highest common factor springs to mind.

5. Nov 30, 2005

### TimNguyen

Oh... that's what I meant. So if (m,n)=1, then only a could factors both a^n and a^m?

6. Nov 30, 2005

### matt grime

That still doesn't make any sense. Don't think in G or S for now.

If I were to say: let m and n be coprime, then what must your first reaction be? (Don't mention G or S at all, please)

7. Nov 30, 2005

### TimNguyen

Then the greatest common divisor for m and n is 1.

8. Nov 30, 2005

### matt grime

Well, obviously, since that is the definition of coprime, but what can you deduce from that (still ignoring G and S)? You have done Euclid's Algorithm?

9. Nov 30, 2005

### TimNguyen

Yes, I've done Euclid's algorithm. So it states that dividing m and n over and over until there's a remainder that cannot be factored again will be the greatest common divisor. But since m and n are relatively prime, then I could find integers, p and q, such that mp + nq = 1?

10. Nov 30, 2005

### matt grime

Yes, and now what can you do with those when we allow ourselves to think about G and S and in particular a?

11. Nov 30, 2005

### TimNguyen

Oh... I think I get it. Am I supposed to think in terms of a^mp + a^nq = a^1?

12. Nov 30, 2005

### matt grime

you can't add elements of the group like that. remember that multiplication of powers of an object adds the indices, ie (x^r)(x^s)=x^(r+s)

13. Nov 30, 2005

### TimNguyen

So... if a^m is in S and a^n is in S, then neccessarily (a^m)(a^n) is in S which is equal to a^(m+n)?

14. Nov 30, 2005

### matt grime

yes, that is certainly true, but if x is in S, then x^p is also in S, isn't it, since S is a subgroup. I haven't picked p at random, and y^q would also be suggestively useful.

15. Nov 30, 2005

### TimNguyen

Wow, thanks for all your help Matt.