Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
13
votes
Examples of finite groups with "good" bijection(s) between conjugacy classes and irreducible...
This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree …
10
votes
When do the sizes of conjugacy classes and squares of degrees of irreps give the same partit...
@Marty Isaacs: There exist non-nilpotent groups whose conjugacy class sizes are all squares. For example, let $G$ be Magma's 93rd group order 540. It has class sizes 1,4,9. Indeed, |G|=15*1+30*4+45*9. …
8
votes
Generalization of Schur's Lemma
Schur's lemma has a different generalization when the coefficient field $F$ is not algebraically closed. Then you get $M_{m_1}(D_1)\times\cdots\times M_{m_k}(D_k)$ where $D_i:={\rm Hom}_{FG}(\rho_i,\r …
8
votes
1
answer
468
views
Finite nilpotent orbits: GL(n,q)-conjugacy classes and a partial order on partitions
I have a question regarding a partial order $<$
on the set ${\rm Part}(n)$ of partitions of $n$.
Given $\lambda=(\lambda_1,\lambda_2,\ldots)\in{\rm Part}(n)$ with
$\sum_{i\geq1} \lambda_i=n$ and $\lam …
6
votes
Are there any known criteria for quadratic mapping from R^n to R^n being surjective?
If $n=2$ the quadratic map $\mathbb{R}^2\to\mathbb{R}^2$ with $(x_1,x_2)\mapsto (x_1^2-x_2^2,2x_1x_2)$ is surjective. This follows because the map $\mathbb{C}\to\mathbb{C}$ with $z\mapsto z^2$ is surj …
3
votes
1
answer
408
views
Must normalizing field outer automorphisms "divide" the dimension?
Imprecise question: To get a normalizing field outer automorphism of
order $r$, must we multiply the dimension by $r$?
Precise hypothesis: Let $p\geqslant 5$ be a prime, let $q$ be a power of $p$ and …
3
votes
Constituents of induced representation
In the case that $n$ is prime and $H$ is normal in $G$. Your second questions has an affirmative answer. The composition length is $1$, $n$, or $1+d$ where $d\mid (n-1)$, in each case at most $n$. See …
2
votes
Reference for restriction of a simple module over a splitting field to a smaller field?
One uses a matrix version of Hilbert's Satz 90 to answer Geoff's comment "If the field $E$ is not this minimal field, it seems less obvious to me how to realise the representation over the subfield ge …
2
votes
Richness of the subgroup structure of p-groups
The comment by Frieder Ladisch suggests to me that considering exponents may be relevant. Suppose that we generalize Stefan Kohl's function
$f_p(n)$ as follows:
Definition: Fix a prime $p$ and an exp …
2
votes
Order of products of elements in symmetric groups
First let me paraphrase the question. Given integers $m,n,k$ each at least 2, set $d:=\max(m,n,k)+2$. Do there exist elements $a,b$ in the symmetric group $S_d$ such that $|a|=m$, $|b|=n$ and $|ab|=k$ …
1
vote
Must normalizing field outer automorphisms "divide" the dimension?
The answer to this question is No.
Let $U$ be the natural module for $H=\textrm{SL}(2,5^5)$ and let
$V=U\otimes U^\sigma\otimes\cdots\otimes U^{\sigma^4}$ where $\sigma$ is
field automorphism $\lambda …
1
vote
How to detect if a subgroup lands inside an orthogonal group?
This answer may be more general than you need. Let $F$ be a field, and let
$G=\langle g_1,\dots,g_m\rangle$ be a subgroup of ${\rm GL}(n,F)$ which preserves a form, with matrix $J$. Then $g_i J g_i^{\ …
1
vote
A finite group that splits and does not split
I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your q …