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For a natural number $n$, let $Z_n=\mathbb{Z} \ast \cdots \ast \mathbb{Z}$ denote the free product of $n$ copies of the integers. Let $m$ be a further integer.

$\textit{Question:}$ Is there a way of counting the subgroups of $Z_n$ with index $m$? If that's not possible, what if we restrict our attention to normal subgroups?

I would already be happy to know a way for counting them for $m=2,3$.

Michael Hardy
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user352941
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1 Answers1

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For $m=2$, our subgroups are automatically normal. Thus we need only count the number of homomorphisms $Z_n\to\Bbb Z/2\Bbb Z$. Each such homomorphism is determined by where it maps the generators of the $n$ factors of $Z_n$, hence there are $2^n$ such homomorphisms. After dropping the trivial homomorphism, we conclude that there are $2^n-1$ subgroups of index $2$.

The same method would tell us that there are $3^n-1$ normal subgroups of index $3$. Could it happen that there are also non-normal subgroups?

Hagen von Eitzen
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  • It should be $(3^n-1)/2$. – Derek Holt Jul 09 '16 at 16:53
  • I'm not sure if it was a rhetorical question, but yes, there are non-normal subgroups of index 3. For example $Z_2$ has $S_4$ as a homomorphic image because it is $2$-generated. Since $S_4$ has a non-normal subgroup of index 3, $Z_2$ does too, by the correspondence theorem. (Also, +1.) – Ben Blum-Smith Jul 09 '16 at 16:59