Let $ G $ be a Lie group and $ H $ a connected subgroup of $ G $. If $ N_G(H)/H $ is finite does that imply $ H $ must be closed in $ G $?

Asked Sep 06 '24 at 13:51

Active Sep 06 '24 at 14:16

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An obvious generalization of this claim is: If $ N_G(H)/H $ is closed in $ G/H $ does that imply $ H $ must be closed in $ G $? That is certainly false since we can take $ G=T^2 $ and $ H $ the line with irrational slope on the torus. Then $ N_G(H)=G $ and so $ N_G(H)/H=G/H $ is certainly closed in $ G/H $ but $ H $ is not closed in $ G $.

Motivation: I'm considering that the question Normalizer of connected subgroup might need the assumption that $ H $ is closed. But since $ N_G(H)/H $ is already assumed to be finite, and $ H $ assumed connected, perhaps that already implies that $ H $ must be closed?

edited Sep 06 '24 at 14:16

asked Sep 06 '24 at 13:51

Ian Gershon Teixeira

10,176

This is clearly false. Maybe it is true for connected subgroups. – Moishe Kohan Sep 06 '24 at 14:04
@MoisheKohan Ah ok let me add that $ H $ is connected, since that is already the assumption in the linked question that motivated this. Out of curiosity do you have any particular counterexample in mind for the case that $ H $ is not connected? – Ian Gershon Teixeira Sep 06 '24 at 14:15
@Barri: Both assertions you've made are false. – Moishe Kohan Sep 06 '24 at 15:28
So they are! I will delete my comment, thanks for checking – Barri Sep 06 '24 at 15:43
2

Yes, my example is $PSL(2,Q)<PSL(2,R)$. – Moishe Kohan Sep 06 '24 at 15:55

Let $ G $ be a Lie group and $ H $ a connected subgroup of $ G $. If $ N_G(H)/H $ is finite does that imply $ H $ must be closed in $ G $?

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