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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" –  DonAntonio Nov 15 '12 at 15:46
    
Hint: What are the simplest non-trivial subgroups? –  Thomas Andrews Nov 15 '12 at 15:48
    
Ah, I get it. Suddenly it turns out to be quite easy. –  Ezra Nov 15 '12 at 15:51

1 Answer 1

up vote 1 down vote accepted

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

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