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If a cyclic subgroup has a proper subgroup of infinite order, how many finite subgroups does it have? Explain. Any help appreciated! Thank you.

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  • $\begingroup$ Cyclic subgroup in what exactly? $\endgroup$
    – kahen
    Nov 28, 2010 at 20:36

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It's easy to see that a cyclic infinite group is isomorphic to the set of integers $\mathbb{Z}$: being cyclic, it is generated my one element, $x$ say, and being infinite, no power of $x$ is equal to the identity. Consequently it is formed by the integer powers of $x$, and not two of them are equal.

The answer to your question consequently is that an infinite cyclic group cannot have a nontrivial finite subgroup.

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