E. Kowalski's blog






         Comments on mathematics, mostly.

03.05.2014

More conjugation shenanigans

Filed under: Exercise,Mathematics @ 16:46

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

1 Comment

  1.   Johan — 04.05.2014 @ 2:25    Reply

    Fun!

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