Let $X_G$ be the number of normal subgroups of a group $G$. Are there examples of finitely generated groups $G$ where it is consistent to have $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups? Also are there examples where $\aleph_0<X_G/\mathord\sim<2^{\aleph_0}$ where $N \sim M \iff G/N \cong G/M$?
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$\begingroup$ Wouldn't such an example produce a counterexample to the Continuum Hypothesis? $\endgroup$– sdsCommented Mar 18, 2014 at 20:44
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5$\begingroup$ @sds I asked about the whether it was consistent for an example to exist, and it is consistent for the continuum hypothesis to be false. $\endgroup$– user35370Commented Mar 18, 2014 at 20:56
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1 Answer
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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).
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3$\begingroup$ I believe the same also holds for the second question. For $G$ finitely generated, $N\sim M$ is a Borel equivalence relation on a Polish space, and as such it has countably many or $2^\omega$ equivalence classes. $\endgroup$ Commented Mar 18, 2014 at 18:20
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2$\begingroup$ @Emil: it's even simpler, since this relation has countable classes. $\endgroup$– YCorCommented Mar 18, 2014 at 21:07
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$\begingroup$ Ah, yes, you are right, I missed that. $\endgroup$ Commented Mar 18, 2014 at 22:21
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3$\begingroup$ However, this argument does not apply to countable infinitely generated groups, as then the classes may have cardinality $2^\omega$. The equivalence relation is analytic in this case, so a priori the number of equivalence classes can be countable, $\aleph_1$, or $2^\omega$. I wonder whether $\aleph_1$ is possible (assuming $\neg$CH, obviously). $\endgroup$ Commented Mar 19, 2014 at 13:46
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5$\begingroup$ That’s right, see e.g. mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0269.0276.ocr.pdf . The original result is due to Burgess ams.org/journals/proc/1978-069-02/S0002-9939-1978-0476524-6 . $\endgroup$ Commented Mar 19, 2014 at 18:44