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S Dec 15 at 20:06 history bounty ended CommunityBot
S Dec 15 at 20:06 history notice removed CommunityBot
Dec 15 at 0:03 comment added Ian Gershon Teixeira ah ok, no worries
Dec 14 at 23:52 comment added Moishe Kohan Sorry, I found a mistake in my example.
Dec 13 at 20:15 comment added Ian Gershon Teixeira @MoisheKohan don't want to rush you but if you post in the next 22 hours I can award you the bounty!
Dec 8 at 23:55 comment added Moishe Kohan OK, I will add an answer later.
Dec 8 at 21:55 comment added Ian Gershon Teixeira @MoisheKohan would you be willing to share the specifics of this example, perhaps posted as an answer if it is too long for a comment?
Dec 8 at 20:40 comment added Moishe Kohan Yes, in SL(2,C).
Dec 8 at 19:59 comment added Ian Gershon Teixeira @MoisheKohan That sounds right, do you have a particular example in mind?
Dec 8 at 18:56 comment added Moishe Kohan It fails to be closed under inverse if you drop the irreducibility assumption.
S Dec 7 at 19:03 history bounty started Ian Gershon Teixeira
S Dec 7 at 19:03 history notice added Ian Gershon Teixeira Authoritative reference needed
Dec 7 at 18:51 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 5 at 20:07 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 5 at 18:59 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 5 at 18:51 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 5 at 17:55 comment added Andrea Marino You are right. Not sure is a good way to prosecute, but to keep things small one could try to address the more general problem of describing which $X$ has the property I mentioned. By the way, do you have any hints about why this should hold? From a non-expert perspective, it seems like a pretty un-motivated property!
Dec 5 at 16:56 comment added Ian Gershon Teixeira @AndreaMarino I believe that the irreducible $ H=S_4 $ subgroup of $ GL(3,\mathbb{C}) $ you describe is essentially self-normalizing so the hierarchy immediately terminates with $ N^r(H)= <H,\mathbb{C}^\times I> $ for all $ r \geq 2 $
Dec 5 at 8:49 comment added Andrea Marino Can we understand a simple case? E.g. take $H= S_4$ acting on $\mathbb{R}^3 \simeq \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4: x_1+x_2+x_3+x_4 = 0\}$ by permuting variables (how's that called? Standard?). I was wondering if one can 'classify' sets $X$ that satisfy $g H g^{-1} \subset X$ iff $g^{-1} H g \subset X$
Dec 4 at 23:05 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 4 at 22:56 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 4 at 22:48 comment added LSpice Re, ah, I see. I was thinking, as in @YCor's now-deleted answer, of the algebraic case, where one can show with finite invariants (dimension and number of components) that what I guess should be called the "semi-normalizer" $N^2(H)$ is actually the normalizer; but now I take your point. For example, one could take $\tilde H$ to be the integer-indexed sequences of reals and $H$ to be the subgroup of $\tilde H$ that vanishes on negative integers, and put $G = \langle\sigma\rangle \ltimes \tilde H$, where $\sigma$ is the right shift.
Dec 4 at 22:43 comment added Ian Gershon Teixeira @LSpice I really do mean $ N^2(H) $ fails to be closed under inverses! The subtle thing here is that there is a containment not an equality in the condition $ gHg^{-1} \subset N^r(H) $ so if $ H $ is infinite then even for $ r=1 $ the containment $ gHg^{-1} \subset H=N^1(H) $ can be proper! So although when $ H $ is finite then $ N^2(H) $ is the normalizer, when $ H $ is infinite then $ N^2(H) $ is just some set that (properly) contains the normalizer.
Dec 4 at 22:31 comment added LSpice I think you mean that $N^3(H)$, not the normaliser $N^2(H)$, might fail to be closed under inverses. Is your revised expectation due to the counterexamples you mention, or some other evidence or hunch?
Dec 4 at 21:08 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 3 at 20:06 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 1 at 0:49 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 1 at 0:47 comment added Ian Gershon Teixeira @LSpice +1 good point, you are kind and wise as always!
Nov 30 at 22:59 comment added LSpice I think the norm (as opposed to the normaliser 😄) is that it is better to avoid editorialising, like "interesting", "difficult", etc., in the title (and preferably in the body), so I edited out "interesting".
Nov 30 at 22:58 history edited LSpice CC BY-SA 4.0
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Nov 30 at 22:49 history edited YCor CC BY-SA 4.0
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Nov 30 at 18:46 history asked Ian Gershon Teixeira CC BY-SA 4.0