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The set of normal subgroups (resp. subgroups) of a countable group $G$ is a closed subset of the Cantor set $2^G$. Hence it is either (at most) countable, or contains a Cantor set and hence has cardinal $2^{\aleph_0}$.

If moreover $G$ is finitely generated (as assumed in the question), then the equivalence relation $\sim$ has (at most) countable classes, and it follows that the number of isomorphism classes of quotients of $G$ is also either at most countable, or has continuum cardinal (in ZFC, regardless of the continuum hypothesis). As noted by Emil, it is natural to wonder whether we can reach the same conclusion when $G$ is only assumed countable (in which case the relation $\sim$ may have uncountable classes).

The set of normal subgroups (resp. subgroups) of a countable group $G$ is a closed subset of the Cantor set $2^G$. Hence it is either (at most) countable, or contains a Cantor set and hence has cardinal $2^{\aleph_0}$.

The set of normal subgroups (resp. subgroups) of a countable group $G$ is a closed subset of the Cantor set $2^G$. Hence it is either (at most) countable, or contains a Cantor set and hence has cardinal $2^{\aleph_0}$.

If moreover $G$ is finitely generated (as assumed in the question), then the equivalence relation $\sim$ has (at most) countable classes, and it follows that the number of isomorphism classes of quotients of $G$ is also either at most countable, or has continuum cardinal (in ZFC, regardless of the continuum hypothesis). As noted by Emil, it is natural to wonder whether we can reach the same conclusion when $G$ is only assumed countable (in which case the relation $\sim$ may have uncountable classes).

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YCor
  • 63.9k
  • 5
  • 187
  • 286

The set of normal subgroups (resp. subgroups) of a countable group $G$ is a closed subset of the Cantor set $2^G$. Hence it is either (at most) countable, or contains a Cantor set and hence has cardinal $2^{\aleph_0}$.