Answer to Question #181722 in Algebra for Pradeep Fauzdar

Question

Prove that every non-trivial subgroup of a cyclic group has finite index. Hence 

prove that (Q, +) is not cyclic.

Answer 1

If M

M is a maximal subgroup, the quotient Q/M

Q/M is a simple abelian group, hence certainly finite (actually of prime order): any of its non identity element is a generator, so the group is cyclic and it cannot be infinite cyclic.

Now the task is to show the stronger statement that Q

Q has no proper subgroup of finite index.

If H

H is a subgroup of finite index n

n, then nQ⊆H

nQ⊆H: indeed, for every q∈Q

q∈Q, n(q+H)=H

n(q+H)=H, which amounts to saying that nq∈H

nq∈H. However, nQ=Q

nQ=Q, so H=Q

H=Q.hence it can not be cyclic

You can n ote that no divisible abelian group has proper subgroups of finite index, with exactly the same proof as for Q

Q.