# The number of maximal subgroups up to isomorphism

Every maximal subgroup of infinite index of a free non-cyclic group $F_k$ is free of countable rank. Thus even though the set of maximal subgroups of $F_k$ is uncountable, there are only countably many isomorphism classes of such subgroups (including subgroups of finite index). The same is true for surface groups of genus at least 2. Many other finitely generated groups have at most countable set of isomorphism classes of maximal subgroups.

Question: Is there a finitely generated group with uncountably many isomorphism classes of maximal subgroups?

## 1 Answer

Yes. Let $F$ be a nonabelian f.g. free group, then $F\times F$ has this property. Indeed, for each normal subgroup $N$ of $F$, denote $H_N$ the subgroup $\{(x,y), xN=yN\}$ of $F\times F$ (this is a fibre product $F\times_{F/N} F$). Then $H_N$ contains the normal subgroup $N\times N$, and the quotient $H_N/(N\times N)$ is isomorphic to $F/N$. This normal subgroup can be intrinsically defined in $H_N$, because it is generated by $N\cup\{1\}\cup \{1\}\times N$, which is the set of elements of $H_N$ whose centralizer in $H_N$ is nonabelian. Hence the isomorphism class of $H_N$ determines the isomorphism class of $F/N$. Moreover, if $F/N$ is simple then $H_N$ is maximal in $F\times F$. Since there are uncountably many non-isomorphic 2-generated simple groups, we deduce that $F\times F$ has uncountably many non-isomorphic maximal subgroups.

Edit: here's a reference for the fact that there exists uncountably many non-isomorphic 2-generated simple groups:

Camm, Ruth. Simple free products. J. London Math. Soc. 28, (1953). 66-76.

[MR review behind paywall: (...) we have a continuum of non-isomorphic simple groups $G(\rho,\sigma,\tau)$. Each of these is a free product, with an amalgamated subgroup, of two free groups, and so is, in particular, locally infinite. Moreover, each can be generated by two elements.]

• This is a theorem of Baumslag-Roseblade, 1984 – Igor Rivin May 12 '16 at 18:47
• @Igor: I do not think they mention maximal subgroups there, but they do prove that F x F contains uncountably many pairwise non-isomorphic subgroups. – Mark Sapir May 12 '16 at 19:51
• The fact that $N\times N$ is maximal if $F/N$ is simple indeed follows from Baumslag and Roseblade. By their tesult every subgroup containing $N\times N$ corresponds to a proper factor of $F/N$. – Mark Sapir Jun 15 at 20:11