cragar
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Homework Statement
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.
The Attempt at a Solution
ok so we know there will be some element a that is in G such that
[itex]a^m=e[/itex] 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. Let's 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 [itex]a^m=e[/itex] so if we have some arbitrary power of a [itex]a^x[/itex]
we want its inverse [itex]a^xa^p=e=a^{x+p}=a^m[/itex] 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
[itex]b^n=e[/itex] and then look at all the possible group operations. I guess I could try to find the generator for G.