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Is there a theorem that can help us count easily the number of subgroups of any given finite group?

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    In general? Nope.2012-10-25
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    Of a finite group? Just brute force it! Of an infinite group? Well, there are infinitely many subgroups (take the cyclic subgroup corresponding to each element)!2012-10-25
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    @user1729: There might be different elements generating the same subgroup, though. Is it clear that every infinite group has infinitely many subgroups? It would be enough to find one element generating an infinite subgroup because that subgroup will be cyclic.2012-10-25
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    @Rasmus: You would then have $a=b^i$ and $a^j=b$, so $a$ and $b$ both have order $ finite. In an infinite group you either have an element of infinite order, or infinitely many elements of finite order. This means that every infinite group has infinitely many subgroups.2012-10-25

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Here is a partial answer by G.A. Miller: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1077761/pdf/pnas01604-0036.pdf

It turns out that there is a way to count the subgroups of any given Abelian group.