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The question has been described in the title. How to prove it?

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    It must be "to each of its proper *non-trivial* subgroups"2012-11-15
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    Hint: What are the simplest non-trivial subgroups?2012-11-15
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    Ah, I get it. Suddenly it turns out to be quite easy.2012-11-15

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Look at a cyclic subgroup generated by any element that is not the identity.