lie groups – Why is the quotient by the Borel subgroup compact?

Let $G$ be a complex lie group with semisimple lie algebra $g$.

Put $b$ the borel subalgebra (so the sum of $h$ and all the positive roots) for some choices.

One can there is the borel subgroup $B$ with lie algebra $b$ in $G$; given as a the stabilizer of $b$ in the adjoint action. My book claims now that $G/B$ is compact, since it’s isomorphic to its image in the Grassmanian (i.e where the adjoint sends $b$) and the Grassmanian is compact. However, how do we know that the image is closed?

Subgroup rank of finite simple groups

Definition: The subgroup rank of a finite group G is the minimal natural number n such that every subgroup of G can be generated by n elements (or fewer).

This invariant has been studied extensively for various families of groups. I am interested in the family of finite simple groups and I have been unable to find and relevant information in the literature.

Question 1: Are there only finitely-many finite simple, non-abelian groups G of a given subgroup rank n?

Some relatively straight-forward comments and reductions:

It is not too difficult to show that there are only finitely-many alternating groups of subgroup rank at most n (by explicitly constructing elementary-abelian subgroups of a certain subgroup rank). There are also only finitely-many sporadic groups, according to the classification. These observations reduce the above question to finite simple groups of Lie-type.

Question 2: Are there only finitely-many finite simple groups G of Lie-type with given subgroup rank n?

It is again not too difficult to show that the “field rank” of G is bounded from above by a function of n (by looking at the natural homomorphism from the field to the root subgroups). It is also possible to show that the Lie-rank of G is bounded from above by a function of n. These observations further reduce question 2 to bounding the defining characteristic of the simple group of Lie-type by some function that depends only on the subgroup rank n. Unfortunately, I do not have any good intuition to determine whether the latter statement is true or not.

I hope both questions have a positive answer because that would give us a nice property about the FSG. But I suspect we can prove the answers to be “no” by simply making some judicious choice for the Lie-type, field-rank, and Lie-rank and by then looking at the structure of the Sylow-subgroups of G, as the characteristic goes through the different primes.

gr.group theory – Computable change in minimum word length of subgroup elements

Let $G$ be an infinite finitely generated group. Fix a finite generating set for $G$.

Define $mathrm{len}_G:Gtomathbb{Z}_{geq 0}$ by sending $g$ to the minimum length of a word in the generators and their inverses equal to $g$.

Let $Hsubset G$ is an infinite finitely generated subgroup. Fix a finite generating set for $H$.

Question. Under what conditions is there a computable function $m colon mathbb{Z}_{geq 0}tomathbb{Z}_{geq 0}$ such that for all $hin
H$
the inequality $$ mathrm{len}_H(h)leq m(mathrm{len}_G(h)) $$
holds?

lie groups – Rank of a Lie subgroup generated by two Lie subgroups

Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By the answer to the question asked here

In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?

$L=langle K, Hrangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=rank~(Hcap K)=rank~H=rank~K.$$ My question is if we can conclude that $$rank~L=r.$$

group theory – An easy example of a non-quasiconvex subgroup

Let $G$ be a finitely generated group, and consider the surjection $mu:F(A)to G$ induced by the set of generators $A$, where $F(A)$ is the free group on $A$. A word $w$ is said to be ($mu$-)geodesic if is it of minimal length in $mu^{-1}mu(w)$.
A subgroup $H<G$ is called quasi-convex if there exist a constant $k$ such that for any geodesic word $w=w_1dots w_n$ such that $mu(w)in H$, and for any $0<i<n$, there exist a word $v_i$ of length at most $k$ such that $mu(w_1dots w_iv_i)in H$.

What is the easiest example of a non-quasi-convex subgroup? Here easy may mean with an explicit presentation, easy to prove that is not quasiconvex, or with a very quick and elegant description, depending on the taste of who is answering.

number theory – What’s the torsion subgroup of $y^2 = x^3+2015^2+2015^3$?

$Q$. (a) Let $E$ be the elliptic curve given by $y^2 = x^3+2015^2+2015^3$. Prove that $E(mathbb Q)_{text{tors}}$ is the trivial group. (b) Now let $E$ be the elliptic curve $y^2 = x^3+x+2$. Determine $E(mathbb Q)_{text{tors}}$.

What’s a good method for part (a)? In the attempt below, at some point I’ve to show that $E(mathbb F_7)_{text{tors}} cong mathbb Z/ 11 mathbb Z$. During this do I’ve to check that every point I find is torsion? Or is there some fact that allows to know the points on my list are indeed torsion without any checking? In addition I can’t finish part (b) after showing $E(mathbb F_5) cong mathbb Z/ 4 mathbb Z cong E(mathbb F_3).$

(a) The discriminat $Delta = -5^6 cdot 13^6 cdot 31^7$. This means $E$ has good reduction for all primes $P ne 2,5,13,31.$ In particular, for ${E}(mathbb F_3):$ $y^2= x^3+x+2$ we have $E(mathbb F_3)_{text{tors}} cong mathbb Z/ 4 mathbb Z, $ as the torsion points in this case are $4(1,1) = 4(1, 2) = 2(2,0) = mathcal{O}$.

Since the next good prime is $7$ we can reduce ${E}(mathbb F_7):$ $y^2= x^3+x+6$, which has points:

$left{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)right} $

Then I find the order of every point on this list and conclude that:

and $left{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)right} cup left{mathcal Oright} cong mathbb Z/ 11 mathbb Z.$

Therefore $E(mathbb F_7)_{text{tors}} cong mathbb Z/ 11 mathbb Z. $ Then the order of $E(mathbb Q)_{text{tors}}$ divides $gcd(4,11) = 1$ therefore $E(mathbb Q)_{text{tors}}$ is the trivial group.

(b) If ${E}/ mathbb Q:$ $y^2= x^3+x+2$ then $E(mathbb F_3) cong mathbb Z/ 4 mathbb Z$ as above, but also similarly $E(mathbb F_5) cong mathbb Z/ 4 mathbb Z$. I can’t finish this but I’m guessing it somehow follows it from part (a). If so, how?

abstract algebra – Subgroup of order n of group of order 2n

There is a subgroup of order $n$ in a group of order $2n$. Prove, that all squares of elements lie in this subgroup.

I thought about some solution, with the usage of Lagrange theorem and normality of this subgroup, but I didn’t come up with anything good.

gr.group theory – Irreducible representation of $U(n)$ as subgroup of $U(d)$

$U(n)$ is the group of $n$-dimensional complex unitary matrices. This group has irreducible representations in higher dimensions. Let $R_{n,d}$ be an irrep of $U(n)$ in terms of $d$-dimensional matrices, $d>n$. The matrices in $R_{n,d}$ are unitary, so this is a subgroup of $U(d)$,
$$R_{n,d}subset U(d).$$

What is known about the way $R_{n,d}$ sits inside $U(d)$? For example, what is $U(d)/R_{n,d}$?

What is an isolated subgroup?

Infinitesimal numbers form an isolated subgroup of finite numbers.
What does this sentence mean? What is an isolated subgroup?

reference request – Finitely-generated conjugation action on a subgroup that is not normal… what is that?

If $H lhd G$, then $G$ acts on $H$ by conjugation. I need to talk about this action but in a situation where $H$ is not (necessarily) normal. When $H leq G$, there is a “partial action” of $G$ on $H$ by $G$-conjugation, and clearly it has very nice properties, for example $h^g h^{g’} = h^{gg’}$ and $(hk)^g = h^g k^g$ when these expressions are defined. To my amazement (at how bad I am at searching), I did not find a discussion of this anywhere. I’m sure it is very standard, and I’m looking for some pointers for a good formal setting.

Question. Is there a good basic reference for the “partial conjugation action” of a group on a subgroup, in some formalism? Is there a standard way to talk about this object?

I know many ways to talk about partial actions, and I can specialize them to my case, so in theory I can solve this question in many ways myself. However, it seems like such a natural example of a partial action that the fact I am not finding this discussed anywhere suggests to me that I may be missing something, so a reference or situation where this appears would be nice.

A more serious problem that tripped me up is “generation”. In many of the settings of partial actions, it is difficult to state that a partial group action is finitely-generated. If $G = langle S rangle$, then the partial actions given by partial conjugation by $s in S$ on a subgroup $H leq G$ obviously “generate” the partial action of $G$ in some sense. But I don’t really know how to say this in a good way, especially I run into issues with domains, when trying to state write down the axioms, and I don’t want to reinvent the wheel.

More specifically, it is natural enough to say that a group action $G curvearrowright H$ is finitely generated when $G$ is, so…

Question. Is there a standard way to say a “partial group action” is “finitely-generated”, so that in the case above of a finitely-generated group $G$ partially acting on its subgroup $H$ by conjugation, the partial action of $G$ would indeed be finitely-generated?

I tried to look at some existing formalisms for partial actions. One thing you can do is form the action groupoid of $G$ acting on itself by conjugation, so take the action groupoid with $Gamma = G$ acting group, $Omega = G$ the set where it acts, then take the subgroupoid for the restriction $H subset Omega$. Unfortunately then it is a bit awkward (I think) to discuss individual partial bijections that may or may not end up being part of such an action (they should of course be “partial automorphisms”, but the list of axioms does not seem to be suggested by the subgroupoid definition). It is also not obvious to me how to state finite generation correctly. Groupoidification drops the group, and after taking the subgroupoid corresponding to $H$, it might not even be determined up to isomorphism, so when you say that a groupoid of partial automorphisms on $H$ is finitely-generated, there is no $G$ this could possibly refer to. Furthermore a “finitely-generated groupoid” seems to usually have a finite number of objects, so this doesn’t look correct.

I then thought of pseudogroups, but all pseudogroup references I found deal with pseudogroups of homeomorphisms, and discuss mostly orthogonal issues. By any definition of a pseudogroup I could find, an action by partial automorphisms is not really a pseudogroup (unless I introduce some topological structure that I’m not going to use). Furthermore, I did not find a discussion of finite generation that tells me what I should do with domains.

There is also literally the notion of a partial action of a group. I thought of this last because I had never actually seen this before (people I know only talk about groupoids), but this was maybe the most promising formalism. I guess I would like to discuss the representations of groups by the “inverse semigroup of partial automorphisms of a group”, but I don’t know the jargon, and at least based on a brief look I did not find a notion of finite generation with the correct properties.