Answer to Question #203246 in Abstract Algebra for Anand

Question

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

prove that (Q, +) is not cyclic.

Answer 1

Let us prove that every non-trivial subgroup of a cyclic group has finite index. If a cyclic group is finite, then all its subgroup finite and have finite index. If a cyclic group is infinite, then it is isomorphic to the additive group of integers. Taking into account that all subgroups of a cyclic group are cyclic, we conclude that every non-trivial subgroup of a cyclic group is of the form $k\mathbb Z$ for $k\in\mathbb Z\setminus\{0\}.$ The quotient group $\mathbb Z/k\mathbb Z=\{[0],[1],\ldots [k-1]\}$ is of order $k$, and hence every non-trivial subgroup of a cyclic group has finite index.

The group $(\mathbb Q, +)$ contains non-trivial subgroup $(\mathbb Z, +)$ of infinite continuum index: $|\mathbb Q/\mathbb Z|=|[0,1)|=c.$ Hence the group $(\mathbb Q, +)$ is not cyclic.