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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. – Benjamin Dickman Oct 25 '12 at 9:12
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)! – user1729 Oct 25 '12 at 9:47
@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. – Rasmus Oct 25 '12 at 10:42
@Rasmus: You would then have $a=b^i$ and $a^j=b$, so $a$ and $b$ both have order $<ij$ 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. – user1729 Oct 25 '12 at 11:08
up vote 2 down vote accepted

Here is a partial answer by G.A. Miller:

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

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