After I wrote my last post on the condition $latex \xi H\xi =H$ in a group, I had a sudden doubt concerning the case in which this arose: there we assume that we have a coset $latex T=\xi H$ such that $latex g^2\in H$ for all $latex g\in T$. I claimed that this implies $latex \xi H \xi =H$, but really the argument I wrote just means that $latex \xi H \xi \subset H$: for all $latex g\in H$, we get $latex \xi g \xi g \in H$, hence $latex \xi g \xi \in H$. But what about the other inclusion?
I had in mind a case where the groups involved are finite, or close enough, so that the reverse inclusion is indeed obviously true. But it is amusing to see that in fact what I wrote is correct in all cases: if $latex H$ is a subgroup of an arbitrary group $latex G$ and $latex \xi\in H$ satisfies $latex \xi H \xi\subset H$, then in fact $latex \xi H \xi = H$: taking the inverse of the inclusion gives $latex \xi^{-1} H\xi^{-1}=\xi^{-1}H^{-1}\xi^{-1}\subset H^{-1}=H.$
I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition $latex \xi H\xi^{-1}\subset H$ is not, in general, equivalent with $latex \xi H\xi^{-1}=H$.
(I found this explained in an exercise in Bourbaki, Algèbre, Chapitre I, p. 134, Exercice 27, or page 146 of the English edition; here is a simple case of the counterexample from that exercise: consider the group $latex G$ of permutations of $latex \mathbf{Z}$; consider the subgroup $latex H$ which is the pointwise stabilizer of $latex \mathbf{N}=\{0,1,2,\ldots\}$, and the element $latex \xi\in G$ which is just
$latex \xi(x)=x+1.$
Then we have $latex \xi H\xi^{-1}\subset H$, because the left-hand side is the pointwise stabilizer of $latex \xi(\mathbf{N})$ which is a subset of $latex \mathbf{N}$. But $latex \xi H\xi^{-1}$ is not equal to $latex H$, because $latex \xi^{-1} H\xi$ is the pointwise stabilizer of $latex \xi^{-1}(\mathbf{N})$, and there are elements in $latex G$ which fix $latex \mathbf{N}$ but not $latex \xi^{-1}(\mathbf{N})=\{-1,0,1,\ldots\}$.)
It seems natural to ask: what exactly is the set $latex X$ of all $latex (a,b)\in\mathbf{Z}^2$ such that $latex \xi^a H\xi^b\subset H$ is equivalent to $latex \xi^a H \xi^b=H$? This set $latex X$ contains the line $latex (n,n)$ for $latex n\in\mathbf{Z}$, and also the coordinates axes $latex (n,0)$ and $latex (0,n)$ (since we then deal with cosets of $latex H$).
In fact, one can determine $latex X$: it is the complement of the line $latex (n,-n)$, for $latex n\not=0$ (which corresponds to conjugation). Indeed, suppose for instance that $latex a+b\geq 1$. Let $latex H$ and $latex \xi$ be such that $latex \xi^a H \xi^b\subset H$.
We can write
$latex -a=(a+b-1)a-a(a+b),\quad\quad -b=(a+b-1)b-b(a+b),$
and
$latex \xi^{-a}H\xi^{-b}=\xi^{(a+b-1)a} (\xi^{a+b})^{-a}H (\xi^{a+b})^{-b} \xi^{(a+b-1)b}.$
Since $latex \xi^{a+b}\in H$, and since $latex a+b-1\geq 0$ is the common exponent of $latex \xi^a$ and $latex \xi^b$ at the two extremities of the right-hand side, it follows that this right-hand side is contained in $latex H$.
A similar argument works for $latex a+b\leq -1$, using
$latex -a=(-a-b-1)a+a(a+b),\quad\quad -b=(-a-b-1)b+b(a+b),$
where again it is crucial that the coefficient $latex (-a-b-1)$ appears on both sides, and that it is $latex \geq 0$.
Since we already know that $latex (1,-1)\notin X$, and in fact that $latex (-n,n)\notin X$ for $latex n\not=0$ (the same setting as the counterexample above works, because the $latex \xi$ we used is an $latex n$-th power for every $latex n\not=0$), we have therefore determined the set $latex X$…
