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Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
8
votes
Accepted
Representations of $\text{SL}(n,\mathbb{F}_p)$ and $\text{Sp}(2n,\mathbb{F}_p)$ whose dimens...
A full classification of such representations (and much more) can be found here:
Prime power degree representations of
quasi-simple groups by Malle and Zalesskii
You can read this paper here. …
3
votes
1
answer
127
views
Projective representations of extensions of $PSL_2(q)$
Let $K=PSL_2(q)$ where $q=p^a$ for some odd prime $p$, and let $G$ be a group such that $G/O(G)\cong K$. (Here $O(G)$ is the largest odd-order normal subgroup of $G$.)
I have a homomorphism $\phi: G\ …
7
votes
Is there a Murnaghan-Nakayama Rule for GL(n,q)?
I presume you're talking about characters over C. In which case Green's classic paper "The characters of the finite general linear groups" is the only thing I know on this subject. He defines a whole …
17
votes
Accepted
Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree ...
Old answer: You know already that the answer is ``yes.'' For a reference, see result 2 of
Rasala, Richard On the minimal degrees of characters of $S_n$. J. Algebra 45 (1977), no. 1, 132–181.
Thi …
2
votes
Representations of orthogonal groups over the field of two elements
You can find a nice, concise introduction in $\S5.4$ of Kleidman and Liebeck's The subgroup structure of the finite classical groups. They focus on the modular theory over $\overline{\mathbb{F}_2}$, b …
4
votes
Choice of bilinear forms for simple groups
I think the key word you need here is reflexive. A bilinear form $\beta$ is reflexive if
$$\beta(x,y)=0 \Longrightarrow \beta(y,x)=0.$$
It's pretty clear that any sensible notion of orthogonality (and …
4
votes
Subgroups of GL_2 over a finite field
Dickson is responsible for the classification of subgroups of $SL_2(\mathbb{F}_q)$ (and once you've got this the subgroups of $GL_2(\mathbb{F}_q)$ are easy). You can find a full proof in Suzuki's "Gro …
5
votes
Accepted
Subgroups and representations of finite groups of Lie type
Explicit values for the minimum degree of a primitive permutation representation of a simple group of Lie type can be found in Table 4 of this paper:
Guest, Simon; Morris, Joy; Praeger, Cheryl E.; Spi …
5
votes
Accepted
Special linear groups contained in symplectic groups
The question is this:
When can $SL(m,q^k)$ be a subgroup of $Sp(2n,q)$ with $mk=2n$?
As you point out, this is possible if $m=2$. There are many particular cases that can be ruled out by order c …
4
votes
Largest permutation group without 2-cycles or 3-cycles
This is really an comment to @Dima's answer, but it's a bit long...
There is a classical result of Jordan in permutation group theory that says the following:
If a primitive group $G$ [on a set o …
5
votes
Accepted
Regular elementary abelian subgroups of primitive permutation groups
This is a bit long for a comment. To answer the question it would be very useful to know what groups can act primitively on a set of order $2^a$. Looking at O'Nan--Scott--Aschbacher, we see that such …
7
votes
Accepted
If all real conjugacy classes are strongly real, then all real irreps are "strongly real"(sy...
The answer is No. My source is Rod Gow's rather wonderful little paper "Real-valued characters and the Schur index" (MSN). Let me quote from the introduction:
It may happen that all elements of a
…
3
votes
2
answers
235
views
Do the irreducible modules of this finite group preserve a tensor product structure?
I am interested in a particular group $G$, where
$$ (A_4\times C_\ell) \lhd G \lhd S_4 \times D_\ell$$
Here, $C_\ell$ is cyclic, $D_\ell$ is dihedral of order $2\ell$, and the two inclusions both have …
20
votes
3
answers
934
views
What did Frobenius prove about $M_{12}$?
I am interested in this paper which I can't read because it's in German:
Frobenius, G., Über die Charaktere der mehrfach transitiven Gruppen., Berl. Ber. 1904, 558-571 (1904). ZBL35.0154.02.
A free …
2
votes
Common basis for permutation matrices
Here's a possible method:
We might as well assume that $A$ is already a permutation matrix -- just conjugate $A$ and $B$ by the same unitary matrix $C$ and relabel.
Now if we are looking to conjugat …