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In recent work Lifting N operads from conjugacy data on homotopical combinatorics / N operads in equivariant homotopy theory, collaborators Scott Balchin, Ethan MacBrough, and I identified a class of groups we call lossless. Here is our definition:

A group G is lossless when for all KHG and gG such that gKg1H, there exists hNG(H) (the normalizer of H in G) such that hKh1=gKg1.

Here are our questions:

Is this class of groups (a) studied elsewhere in the literature (presumably under a different name)? or (b) specified by some more familiar string of conditions/adjectives?

For our purposes, we would be happy to restrict to G finite.

We already know (see §2.2 of the paper) that finite solvable T-groups, p-groups of order at most p3, and groups with cyclic normal subgroups of prime index are lossless. The group SL2(F7) is lossy (i.e. not lossless) and we give a class of p-groups of order p4 which are lossy in Example 2.18.

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    This is somewhat related to "control of fusion", I think. Sep 21, 2022 at 17:20
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    @SamHopkins - Good catch! I have fixed a typo in the definition of lossless. When gKg1H for some gG, we in fact require hNG(H) such that gKg1=hKh1.
    – kyleormsby
    Sep 21, 2022 at 17:52
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    In finite group theory, ( and transfer, etc), often control of fusion can be setwise. So, here a finite group theorist might say that for each subgroup H of G the fusion of subgroups of H within G is controlled by NG(H). It is true that in work of Alperin and others, and later work on fusion systems, we need control of strong fusion (I've always felt that strong control of fusion would be a better expression), wher we require h and g ( as in your definition) to induce the same conjugation elementwise on K , that is, h1gCG(K)). Sep 21, 2022 at 19:43
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    Often in finite group theory, it is how the fusion of subgroups ( or sometimes just individual elements) of a fixed Sylow p-subgroup S of G is effected within G which is important. In the more genral setting of dealing with all subgroups, as you seem to, then paper of people like A. Dress on idempotents in the Burnside ring of a finite group may be relevant to you (and there have been much work on uses of the Burnside ring in group representation theory over the decades since) Sep 21, 2022 at 20:13
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    Call P(H,K) the given condition (so "lossless" means that P(H,K) holds for all KHG). Then P(H,K) is trivial when H is normal. Note that the condition P(H,H) is trivial when H is finite but fails when there exists gG such that gHg1<H. The existence of a subgroup conjugate to a proper subgroup of itself is quite frequent for infinite groups (e.g. every group with a nonabelian free subgroup, and many others). Anyway, a f.g. group in which every proper subgroup is cyclic, or in which every proper subgroup is finite (cf. Tarski monsters) is lossless.
    – YCor
    Sep 21, 2022 at 21:39

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