# E. Kowalski's blog

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 $X$

01.05.2014

### Normalizers everywhere

Filed under: Exercise,Mathematics @ 13:05

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 $G$.

25.03.2014

### All Hail the distinguished achievement professor!

Filed under: Mathematics @ 13:55

Mr. Quomodocumque is probably too modest to mention it himself, so let me be the first mathematics blogger to congratulate Jordan Ellenberg on becoming a Vilas Distinguished Achievement Professor! Which hopefully comes with a lot of free time to visit Switzerland…

23.03.2014

### On diplomats

Filed under: Arts,France,Literature @ 20:52

Quizz: who wrote

Il a publié il y a deux ans (…) un ouvrage relatif au sentiment de l’Infini sur la rive occidentale du lac Victoria-Nyanza et cette année un opuscule moins important, mais conduit d’une plume alerte, parfois même acérée, sur le fusil à répétition dans l’armée bulgare, qui l’ont mis tout à fait hors de pair.

or, in translation:

He has published two years ago (…) a book concerning the feeling of Infinity on the occidental shore of the Victoria-Nyanza lake, and this year another booklet, less important but written with a lively, and even piercing, pen, on the repeating rifle in the bulgarian army, which have made him rather peerless.

This reminds of the curiously little-known hilarious “Antrobus stories” of diplomatic mishaps:

It was during one of those long unaccountable huffs between ourselves and the Italians. You know the obscure vendettas which break out between Missions? Often they linger on long after the people who threw the first knife have been posted away. I have no idea how this huff arose. I simply inherited it from bygone dips whose bones were now dust. It was in full swing when I arrived — everyone applying freezing-mixture to the Italians and getting the Retort Direct in exchange. (…) So while bows were still exchanged for protocol reasons they were only, so to speak, from above the waist. A mere contortion of the dickey, if you take me, as a tribute to manners. A slight Inclination, accompanied by a moue. Savage work, old lad, savage work!

(from “The game’s the thing”, where a soccer game between the English and Italian embassies rather degenerates.)

01.03.2014

### More things of the day(s)

(1) Today’s Word of The Day in the OED: afanc, which we learn is

In Welsh mythology: an aquatic monster. Also: an otter or beaver identified as such a monster.

Maybe the Welsh otters, like their rugbymen, are particularly fierce?

(2) Yesterday’s Google doodle, in Switzerland at least, celebrated the 57th birthday of Gaston Lagaffe

I’ve heard that Gaston

Billard

is mostly unknown to the US or English, leaving many people with no reaction to the mention of the contrats de Mesmaeker

Contrats

or to the interjection Rogntudju!.

This lack of enlightenment is a clear illustration of the superiority of the continental mind.

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