Thread: Cyclic subgroup

Suppose that the order of some finite Abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

My proof so far:

Suppose G is a finite Abelian group with 10 | |G|. By a theorem, I know that G contains a subgroup with order 10. Now, a subgroup of a Abelian group is also Abelian (is that right? I recall that from an exercise I did), and since G can be written as both of these groups are cyclic, so G contains a cyclic subgroup.

Is that right?

Thanks.

Originally Posted by tttcomrader

Suppose that the order of some finite Abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

My proof so far:

Suppose G is a finite Abelian group with 10 | |G|. By a theorem, I know that G contains a subgroup with order 10. Now, a subgroup of a Abelian group is also Abelian (is that right? I recall that from an exercise I did), and since G can be written as both of these groups are cyclic, so G contains a cyclic subgroup.

Is that right?

Thanks.

Since you are allowed to use the fundamental theorem for finite abelian groups it means the converse of the theorem of Lagrange holds. Thus, there is a subgroup of order 10. Since 10 is square free it means by the fundamental theorem this group is cyclic. Q.E.D.

Originally Posted by tttcomrader

Suppose that the order of some finite Abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

My proof so far:

Suppose G is a finite Abelian group with 10 | |G|. By a theorem, I know that G contains a subgroup with order 10. Now, a subgroup of a Abelian group is also Abelian (is that right? I recall that from an exercise I did), and since G can be written as both of these groups are cyclic, so G contains a cyclic subgroup.

Is that right?

Thanks.

Use Lagrange's theorem ( or the corollary) Ü