Normalizers everywhere

In working on a paper, I found myself in the amusing but unusual situation of having a group G, a subgroup H and an element \xi\in G such that
\xi H\xi =H.

This certainly can happen: the two obvious cases are when H=G, or when \xi is an involution that happens to be in the normalizer N(H) of H.

In fact the general case is just a tweak of this last case: we have \xi H\xi=H if and only if \xi^2\in H and \xi\in N(H), or in other words, if \xi belongs to the normalizer, and is an involution modulo H.

This is of course easy to check. I then asked myself: what about the possibility that
\xi^a H\xi^b = H,
where a and b are arbitrary integers? Can one classify when this happens? The answer is another simple exercise (that I will probably use when I teach Algeba I next semester): this is the case if and only if \xi^{a+b}\in H and \xi^{(a,b)}\in N(H), where (a,b) is the gcd of a and b. In particular, for all pairs (a,b) where a and b are coprime, the condition above implies that \xi belongs to the normalizer of H.

Here is the brief argument: having fixed \xi, let
M=\{(\alpha,\beta)\in\mathbf{Z}^2\,\mid\, \xi^{\alpha}H\xi^{\beta} =H\}.
This set is easily seen to be a subgroup of \mathbf{Z}^2. Furthermore, note that (\alpha,\beta)\in M implies that \xi^{\alpha+\beta}\in H, which in turns means that (\alpha+\beta,0)\in M and (0,\alpha+\beta)\in M.

Hence if (a,b)\in M, we get
(a,-a)=(a,b)-(0,a+b)\in M,\quad\quad (b,-b)=(a+b,0)-(a,b)\in M,
so that
M\cap (1,-1)\mathbf{Z}
contains (a,-a)\mathbf{Z}\cap (b,-b)\mathbf{Z}, which is just
(d,-d)\mathbf{Z},
where d=(a,b). But (d,-d)\in M means exactly that \xi^{d}\in N(H).

Thus we have got the first implication. Conversely, the conclusion means exactly that
(a+b,0)\in M,\quad (d,-d)\in M.
But then
(a,b)=(a+b,0)-(b,-b)=(a+b,0)-b/d (d,-d)\in M
shows that \xi^a H\xi^b=H.

To finish, how did I get to this situation? This can arise quite naturally as follows: one has a collection X of representations \rho of a fixed group \Gamma, and an action of a group latex G on these representations X (action up to isomorphism really).
For a given representation \rho_0, we can then define a group
H=\{g\in G\,\mid\, g\cdot \rho_0\simeq \rho_0\}
and also a subset
T=\{g\in G\,\mid\, g\cdot \rho_0\simeq D(\rho_0)\},
where D(\cdot) denotes the contragredient representation.

It can be that T is empty, but let us assume it is not. Then T has two properties: (1) it is a coset of H (because H acts on T simply transitively); (2) we have g^2\in H for all g\in T (because the contragredient of the contragredient is the representation itself).

This means that, for some \xi\in G, we have T=\xi H, and furthermore
\xi H\xi=H
since \xi g\xi g\in H for all g\in H.

By the previous discussion, we therefore get a good handle on the structure of T: either it is empty, or it is of the form \xi H for some \xi\in G such that \xi^2\in H and \xi normalizes H in G. In particular, if H is trivial (which happens often), either T is empty, or it consists of a single element \xi which is an involution of G.

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Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

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