Let G be an infinite group and let H be a finite subgroup. Does there exist a finite index normal subgroup of G which contains H?

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deliria1t4 · Accepted Answer

Let G be the set of complex numbers which eventually map to 1 under repeated squaring. Then G is a group under complex multiplication.   Let H⊆G be a proper subgroup. As odd numbers are invertible modulo 2n, for n≥1, we have:  e2πi2m+12n∈H⇒e2πi12n∈H.  Thus H is completely characterised by the maximal n∈N such that e2πi12n∈H. Then H is cyclic of order 2n.  In particular H may be finite, but there are no finite index proper subgroups for it to be contained in.

Let G be an infinite group and let H be

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