Subgroup of given order of an Abelian group

[siddhuiitb]

Hey!
We know that if there exists an element of a given order in a group, there also exists a cyclic subgroup of that order. What about converse?
Suppose there is a subgroup of an Abelian group of order 'm'. Does that imply there also exists an element of order 'm' in the Group. It does not hold in general for Non-Abelian groups. But what about Abelian groups?

 

[Hurkyl]

Suppose that were true. what if m = |G|?

 

[lavinia]

Hey!
We know that if there exists an element of a given order in a group, there also exists a cyclic subgroup of that order. What about converse?
Suppose there is a subgroup of an Abelian group of order 'm'. Does that imply there also exists an element of order 'm' in the Group. It does not hold in general for Non-Abelian groups. But what about Abelian groups?

Take a look at an elementary Abelian 2 -group of order greater than 2 e.g. Z/2 x Z/2 x Z/2.

 

[siddhuiitb]

Thanks!

 

[blue_raver22]

Hello!

Yes, in the case of Abelian groups, the converse is also true. If there exists a subgroup of order 'm', then there also exists an element of order 'm' in the group. This is because in Abelian groups, all subgroups are normal, meaning that the subgroup and its cosets form an equivalence class. Since the subgroup has order 'm', each coset will have 'm' elements, and therefore, there will be an element of order 'm' in the group. This is a result of the fundamental theorem of Abelian groups. So, in summary, for Abelian groups, the existence of a subgroup of a given order implies the existence of an element of that order in the group.