Normalizers everywhere – E. Kowalski's blog

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