Question #203246

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

prove that (Q, +) is not cyclic.


1
Expert's answer
2021-06-07T05:03:27-0400

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 kZk\mathbb Z for kZ{0}.k\in\mathbb Z\setminus\{0\}. The quotient group Z/kZ={[0],[1],[k1]}\mathbb Z/k\mathbb Z=\{[0],[1],\ldots [k-1]\} is of order kk, and hence every non-trivial subgroup of a cyclic group has finite index.


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



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