More conjugation shenanigans

After I wrote my last post on the condition $\xi H\xi =H$ in a group, I had a sudden doubt concerning the case in which this arose: there we assume that we have a coset $T=\xi H$ such that $g^2\in H$ for all $g\in T$ . I claimed that this implies $\xi H \xi =H$ , but really the argument I wrote just means that $\xi H \xi \subset H$ : for all $g\in H$ , we get $\xi g \xi g \in H$ , hence $\xi g \xi \in H$ . 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 $H$ is a subgroup of an arbitrary group $G$ and $\xi\in H$ satisfies $\xi H \xi\subset H$ , then in fact $\xi H \xi = H$ : taking the inverse of the inclusion gives $\xi^{-1} H\xi^{-1}=\xi^{-1}H^{-1}\xi^{-1}\subset H^{-1}=H.$

I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition $\xi H\xi^{-1}\subset H$ is not, in general, equivalent with $\xi H\xi^{-1}=H$ .

(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 $G$ of permutations of $\mathbf{Z}$ ; consider the subgroup $H$ which is the pointwise stabilizer of $\mathbf{N}=\{0,1,2,\ldots\}$ , and the element $\xi\in G$ which is just
$\xi(x)=x+1.$

Then we have $\xi H\xi^{-1}\subset H$ , because the left-hand side is the pointwise stabilizer of $\xi(\mathbf{N})$ which is a subset of $\mathbf{N}$ . But $\xi H\xi^{-1}$ is not equal to $H$ , because $\xi^{-1} H\xi$ is the pointwise stabilizer of $\xi^{-1}(\mathbf{N})$ , and there are elements in $G$ which fix $\mathbf{N}$ but not $\xi^{-1}(\mathbf{N})=\{-1,0,1,\ldots\}$ .)

It seems natural to ask: what exactly is the set $X$ of all $(a,b)\in\mathbf{Z}^2$ such that $\xi^a H\xi^b\subset H$ is equivalent to $\xi^a H \xi^b=H$ ? This set $X$ contains the line $(n,n)$ for $n\in\mathbf{Z}$ , and also the coordinates axes $(n,0)$ and $(0,n)$ (since we then deal with cosets of $H$ ).

In fact, one can determine $X$ : it is the complement of the line $(n,-n)$ , for $n\not=0$ (which corresponds to conjugation). Indeed, suppose for instance that $a+b\geq 1$ . Let $H$ and $\xi$ be such that $\xi^a H \xi^b\subset H$ .

We can write
$-a=(a+b-1)a-a(a+b),\quad\quad -b=(a+b-1)b-b(a+b),$
and
$\xi^{-a}H\xi^{-b}=\xi^{(a+b-1)a} (\xi^{a+b})^{-a}H (\xi^{a+b})^{-b} \xi^{(a+b-1)b}.$
Since $\xi^{a+b}\in H$ , and since $a+b-1\geq 0$ is the common exponent of $\xi^a$ and $\xi^b$ at the two extremities of the right-hand side, it follows that this right-hand side is contained in $H$ .

A similar argument works for $a+b\leq -1$ , using
$-a=(-a-b-1)a+a(a+b),\quad\quad -b=(-a-b-1)b+b(a+b),$
where again it is crucial that the coefficient $(-a-b-1)$ appears on both sides, and that it is $\geq 0$ .

Since we already know that $(1,-1)\notin X$ , and in fact that $(-n,n)\notin X$ for $n\not=0$ (the same setting as the counterexample above works, because the $\xi$ we used is an $n$ -th power for every $n\not=0$ ), we have therefore determined the set $X$ …

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Kowalski

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