1. Apr 8, 2013

### cragar

1. The problem statement, all variables and given/known data
Let G be an ableian group of order mn, where m and n are relativiely prime. If G has
has an element of order m and an element of order n, G is cyclic.
3. The attempt at a solution
ok so we know there will be some element a that is in G such that
$a^m=e$ where e is the identity element. It seems that this would be enough to prove that their is a sub group generated by a. and this sub group is cyclic. if I start with the element a
all powers of a would need to be in their so it would be closed under the operation.
I guess we know its a group already. Lets say we have some power of a like x where
0<x<m we want to know if this has an inverse that is a power of a.
we know $a^m=e$ so if we have some arbitrary power of a $a^x$
we want its inverse $a^xa^p=e=a^{x+p}=a^m$ so x+p=m so their is a cyclic subgroup
generated by a, Now we know that if we have a cyclic group all of its subgroups are cyclic.
I am slightly worried about the converse, is it always true if I have cyclic subgroup that the group is cyclic? But I guess i could just do the same argument with some element of the form
$b^n=e$ and then look at all the possible group operations. I guess I could try to find the generator for G.

2. Apr 8, 2013

### jambaugh

What is the order of ab?

3. Apr 8, 2013

### cragar

mn, so I guess ab would be the generator of the group.

Last edited: Apr 9, 2013
4. Apr 9, 2013

### jambaugh

Well, you must argue that. Given its order and the size of the group, what's left?