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Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
7
votes
Accepted
Wedderburn decomposition of special linear groups
$\DeclareMathOperator\M{M}\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\End{End}$As has been mentioned in the comments, the question for algebraically closed fields of characteristic $0$ is equiva …
16
votes
Accepted
Finite groups with integral character table
There is no complete classification, but some structural results are known. To give you something to search for: such groups are called $\mathbb{Q}$-groups. There is a whole book devoted to their stru …
8
votes
Accepted
Irreducible representations of product of profinite groups
This is not even true for finite groups, in this generality, and not even in characteristic $0$. Consider, for example, the group $Q_8 \times C_3$, where $Q_8$ is the quaternion group and $C_3$ is cyc …
1
vote
Reference for fact about reduction mod $p$ of a representation of a finite group
I don't know of just a citation, but here is a pretty quick way to deduce this from the literature (I imagine that this might be the argument you had in mind, in which case apologies for telling you t …
4
votes
Accepted
injective hull and projective cover of simple modules are indecomposable
One definition of "projective cover" of $S$ is that it is a projective module $P$, together with an epimorphism $\phi\colon P\to S$ such that the kernel $K$ is a superfluous submodule of $P$, meaning …
2
votes
Finding all real representations of $\mathrm{SL}_n(\mathbb{F}_q)$
$\DeclareMathOperator{\SL}{SL}\DeclareMathOperator{\GL}{GL}$To determine the real representations of a finite group, it suffices to determine the complex irreducible representations and their Schur in …
16
votes
Accepted
A finite group $G$ all of whose reps are defined over $\mathbb{Z}$ and yet $Rep(G)$ is not g...
The answer is no. Counterexamples include the Weyl groups of types E6, E7, and E8. For a proof that the representations of these Weyl groups are indeed all realisable over $\mathbb{Q}$ (equivalently o …
10
votes
Accepted
For which finite groups $G$ is every character a virtual permutation character?
No classification of such groups is known. As you say, for every character to be a virtual permutation character, necessary conditions are that
all irreducible characters are $\mathbb{Q}$-valued; eq …
6
votes
0
answers
163
views
Generalisation of the Witt–Berman induction theorem
$\DeclareMathOperator{\Aut}{Aut}\DeclareMathOperator{\Ind}{Ind}$I believe I can prove the following induction theorem (modulo carefully checking a few details), and I would like to know whether this i …
1
vote
Schur index of a representation and its divisors
The following is wrong, see comment section:
The Schur index over $K$ is, among other things, the degree of a minimal field extension of $K$ over which the underlying representation can be realised o …
6
votes
Is a finite group given by its character table if its Sylow subgroups are so?
The answer to the first question is negative. The group ${\rm SL}_2(\mathbb{F}_3)$ has a $2$-Sylow subgroup isomorphic to $Q_8$, which is not determined by its character table, but ${\rm SL}_2(\mathbb …
4
votes
Accepted
Even Counterexample to Statement About the Non Existence of Certain Groups with Two Irreduci...
$\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\PGL}{PGL}
\DeclareMathOperator{\mcd}{mcd}
\newcommand{\C}{\mathbb{C}}$Such groups do not exist. Indeed, suppose that $G$ has even order and satisfi …
9
votes
Accepted
How to construct groups and large dimension representations? How about faithful ones?
In the infinite family $G_p=(\mathbb{Z}/p\mathbb{Z})\rtimes (\mathbb{Z}/p\mathbb{Z})^{\times}$, as $p$ runs over prime numbers, the bound is tight up to $O(1)$, and the representation in question is f …
3
votes
Accepted
Local factors determine Weil representations - proof of the cyclic case
Why is the map ${\rm Gal}(FL/L)\to {\rm Gal}(F/K)$ injective?
Elements of ${\rm Gal}(FL/L)$ are automorphisms of $FL$ that act trivially on $L$. To be in the kernel of the above map means to also act …
13
votes
Conjugacy classes of PGL(3,Z)
I will work in ${\rm GL}$ instead of ${\rm PGL}$.
The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of r …