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I do not know where this question is on the trivial to intractable spectrum.

Consider the set of isomorphism classes of groups finitely generated by elements of finite order. What is the cardinality of this set?

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    $\begingroup$ Most likely, continuum, for the same reason that there is continuum of finitely generated groups. $\endgroup$
    – Moishe Kohan
    Commented Apr 15, 2021 at 23:56
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    $\begingroup$ My guess is that every f.g. group embeds into a group generated by finitely many torsion elements. Since each such group contains countably many f.g. subgroups but there are uncountably many f.g. groups, this would imply that there are uncountably many groups generated by finitely many torsion elements. $\endgroup$
    – Andy Putman
    Commented Apr 16, 2021 at 0:01
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    $\begingroup$ Is it possible to compute the cardinality of the set of groups generated by $s,t$ with $s^2=t^3=1$? $\endgroup$
    – Gerry Myerson
    Commented Apr 16, 2021 at 0:12
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    $\begingroup$ Grigorchuk constructed uncountably many groups of intermediate Growth generated by 3 involutions that I believe form uncountably many isomorphism classes but I have to double check $\endgroup$
    – Benjamin Steinberg
    Commented Apr 16, 2021 at 0:55
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    $\begingroup$ @AndyPutman yes: every virtually (free non-abelian) group is SQ-universal. In particular this applies to $\mathrm{PSL}_2(\mathbf{Z})=\langle s,t\mid s^2=t^3=1\rangle$. It's even easier to directly show the latter has continuum many normal subgroups (given that the isomorphism relation between quotients of a given finitely generated group has countable casses, it implies continuum many isomorphism classes). $\endgroup$
    – YCor
    Commented Apr 16, 2021 at 6:36

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Grigorchuk showed that there are uncountably many growth degrees of 2-generated infinite p-groups for any prime p. Hence there are uncountably many isomorphisms classes and even quasi-isometry classes of finitely generated torsion groups. See page 116 of Grigorchuk - On the Gap Conjecture concerning group growth for a discussion. This is of course much stronger than generated by finite order.

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The cardinality of the set of groups generated by three elements of order two is equal to the cardinality of the set of countable groups, i.e., the cardinality of the continuum.

Every countable group embeds in a 2-generator group by the Higman-Neumann-Neumann embedding theorem, and the two generators can be chosen to have infinite order.

Since a 2-generator free group embeds into the free product of three cyclic groups of order 2, it is easy to show (starting from the HNN-embedding theorem) that every countable group embeds in a group generated by three elements of order two.

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    $\begingroup$ The last paragraph needs some elaboration. Indeed if $G\subset F_2/N$ and $F_2\subset H$ with $H$ free product of three cyclic groups of order 2 ($F_2$ has index 2 in $H$), $N$ is possibly not normal in $H$. $\endgroup$
    – YCor
    Commented Apr 16, 2021 at 18:10
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    $\begingroup$ point 37, p.66 of Topics in Geometric Group Theory By Pierre de la Harpe tinyurl.com/ahzwd8ph suggests we only need $\mathbb{Z}_2\ast \mathbb{Z}_3$. $\endgroup$
    – JP McCarthy
    Commented Apr 17, 2021 at 7:21
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    $\begingroup$ Yes, either $\mathbb{Z}_2*\mathbb{Z}_2*\mathbb{Z}_2$ or $\mathbb{Z}_2*\mathbb{Z}_n$ for any $n>2$ will do. $\endgroup$
    – IJL
    Commented Apr 18, 2021 at 10:49

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