1. ## orders of groups

Let x and y be elements of a group G and m and n coprime positive integers. Show that if x^m = y^m and x^n = y^n then x=y

Let x and y be elements of a group G of orders m and n respectively. Show that if x and y commute and m and n are coprime, then order(xy) = mn

not sure at all on this

many thanks

2. If m and n are coprime then there exist integers x and y such that

$\displaystyle xm+yn = 1$

Can you see where to take this?

3. not really.. could you explain a bit more please?

4. Oh, I choose poor letters for that, I meant there are integers a and b such that

am + bn = 1

5. ok.. so we can raise x and y to each of those expressions...

so x^(am).x^(bn) = x
y^(am).y^(bn)=y

since x^m = y^m, x^am = y^am..

so y^am.x^bn = x
hence x = y

or am I waayyyy off track?

thankyou!

6. Nope; nail on the head

7. thankyou!
Any ideas on the next bit (I tried to do it without being able to do the first part - just by assuming the result but still couldnt get anywhere with it.. we've not covered lagrange's theorem)

8. Can you show that (xy)^(mn) = 1?

Can you show that there is nothing smaller which satisfies?

9. I can show that, yes.. to show nothing smaller satisfies would it be sufficient to say, suppose there exists ab < mn such that

xy^ab = 1

then x^ab.y^ab = 1

=> x has order ab.. contradiction?

10. Nope. y^ab could be $\displaystyle (x^{ab})^{-1}$

of a Group G and a proper subgroup H of G such that:
1) G is infinite and H is finite
2) G is infinite and non-abelian and H is abelian
3) G is infinite and abelian and H is infinite
4) G is infinite and H is not-abelian..

I'm really struggling..

for 1 could we have

G = the set of rotations about 0, H = the set of rotations of 90 degress for example?
and simililarly could we have that for 4?

12. ahh ok.. so is contradiction the right way to go about it?

13. ## Some solutions

I dunno if you can use trivial subgroups or not, but that makes some really easy.

1) $\displaystyle G=\mathbb{Z}$ and $\displaystyle H={0}=e$
Another one that isn't as cheap could be polynomials under addition of a finite field, like $\displaystyle Z_p$ and just using H as the subgroup of polynomials with of degree 0 (the constant terms in the field).

2)$\displaystyle G=GL(n,\mathbb{R})$ invertible n by n matrices with entries from the reals under multiplication. If you want to be lame, let H be the identity matrix. Let H be the set matrices that have the same real number on the diagonal and 0s elsewhere. $\displaystyle H={A \in G|A=rI_n, r \in \mathbb{R}}$ Note I_n is the identity matrix.

3)This one is pretty easy, tons of examples $\displaystyle \mathbb{C} \supset \mathbb{R}\supset \mathbb{Q} \supset \mathbb{Z}$ All of these groups are infinite and abelian and the containment is proper, take your pick.

4) You just need some infinite nonableian groups. $\displaystyle G=GL(n,\mathbb{R})$ would be a good start. Then like for subgroups upper triangular matrices might be a good one.

Well that is a start, you should try to think about some more on your own for sure, but this is the idea.