User joseph - MathOverflow

Irreducibility of Induced Representation over arbitrary field

2011-12-29T12:29:21Z

In representation theory of finite groups, there is Mackey test for irreducibility of an induced representation (Serre- Linear Representations of Finite Groups - #7.4). The author has stated it for field $\mathbb{C}$. Is it true for representations over arbitrary field whose characteristic is zero or co-prime to the order of group ?

Uniqueness of splitting field for linear representations of finite groups

2011-10-11T13:09:21Z

If $F$ is any field, $\bar{F}$ its algebraic closure, then it is well-known that all irreducible (indecomposable) $\bar{F}$-representations of a finite group $G$ are realizable over some finite extension $E$ of $F$, and we call $E$, a splitting field for $G$. Any extension of $E$ is also a splitting field of $G$. By finiteness of $[E\colon F]$, we can have a minimal extension of $F$ which is splitting field for $G$.

I puzzled by the following question:

Is such minimal extension unique?

(For example, consider a finite group $G$, $\mathbb{F}_p$ a field of order $p$ with $(|G|,p)=1$. Then there is $n\in \mathbb{N}$ such that $\mathbb{F}_q$ ($q=p^n$) is a splitting field field of $G$.

Is it possible that there are subfields of $\mathbb{F}_q$ of order $p^{m_1}$ and $p^{m_2}$, $m_1\nmid m_2$, and $m_2\nmid m_1$ such that they are splitting fields of $G$, but their subfields are not splitting fields?)

Coverings of a graph of groups

2011-09-05T06:50:51Z

For topological space $X$ (connected, path connected etc.), there is classification of coverings of $X$ : for fixed $x_0\in X$, consider $\pi_1(X,x_0)$. Then there is a $1-1$ correspondance between conjugacy class of subgroups of $\pi_1(X,x_0)$ and covering spaces of $X$ (upto isomorphism). We define universal cover of $X$ to be a cover $\tilde{X}$ which is a cover of every cover $Y$ of $X$. This exist and unique up to isomorphism.

Now consider a graph of groups $(X,A)$, where $X$ is a graph and $A=(A_v,A_e)$ is a family of groups attached to vertices $v\in V(X)$ and edges $e\in E(X)$ of $X$ with injections from edge groups to end-vertex groups.

In the book "Trees"-Serre, the universal cover of $(X,A)$ is defined to be a connected graph $\tilde{X}$ with:

1) a morphism $p\colon \tilde{X}\rightarrow X$ of graphs;

2) an action of $\pi_1(X,x_0)$, $(x_0\in V(X))$ on $\tilde{X}$ such that stabilizer of $\tilde{v}\in p^{-1}(v)$ is isomorphic to vertex group $A_v$, $v\in V(X)$

(In other words, it is a graph, with action of $\pi_1(X,x_0)$, $(x_0\in V(X)$, such the quotient graph of groups $X/\pi_1(X,x_0)$ is isomorphic to given graph of groups).

Question: Is there a construction of universal cover of a graph of groups analogous to the construction of universal cover of topological spaces ( i.e. a cover of every cover of given graph of groups).

As an illustration, how can we obtain all coverings of $(X,A)$ where $X$ is the graph $\circ --\circ$ with vertex groups $\mathbb{Z}/l$, and $\mathbb{Z}/n$ and edge group trivial.

Number of Subgroups of p-Groups

2011-09-09T09:56:56Z

If $G$ is a finite $p$-group of order $p^n$, then it is well known that for ($1\leq m\leq n$), number of subgroups of order $p^m$ is $1$(mod $ p$).

Question: Is it true that number of subgroups of order $p^m$, which are isomorphic within themselves, is $0$(mod $ p$) or $1$(mod $p$).

It looks to be true for groups of order $p^2$, $p^3$.

Embedding of Finite 2-groups

2011-09-07T05:16:47Z

Consider the finite 2-groups containing cyclic subgroup of index 2:

$C_{2^n}$, $C_{2^{n-1}}\times C_2$, $D_{2^n}$, $SD_{2^n}$, $QD_{2^n}$, $Q_{2^n}$.

Can every finite (non-abelian) 2-group be embedded in $H\times K$ where $H$, and $K$ are one of the groups in the list?

Finite quotients of graphs

2011-08-30T02:36:09Z

If $X$ is an infinite graph, $G$ is a group acting on $X$ with finite quotient; make $Y=X/G$ into graph of groups by attaching stabilizers at vertices and edges. Let $Z$ be a graph of groups, with graph equal to graph of $X/G$, all groups at vertices and edged being finite(but not all trivial), and suppose there is morphism of graph of groups $\phi:X/G\rightarrow Z$ , such that $\phi$ is graph isomorphism, and its restriction to vertex groups (and edge groups) is surjective homomorphism.

Does there exist a group whose action on $X$ gives the graph of groups $Z$.

Groups acting on graph

2011-08-29T05:18:05Z

Let $S$ be an infinite graph, $G$ is a group acting (effectively) on $S$ with finite quotient graph $S/G$. Make $S/G$ into graph of groups in obvious way by assigning stabilizers at vertices and edges.

Let $\tilde{S}$ be universal cover of $S$ and $H$ be a group acting (effectively) on $\tilde{S}$ with same quotient as graph $S/G$, but the graph of group may be different.

Question: Does there exist a subgroup $K$ of $G$ which acts on $S$, with quotient graph of groups $S/K$ and $\tilde{S}/H$ isomorphic?

[Here I am failing to use covering space theory directly, because here I am considering quotients spaces with some algebraic structures on them, namely graph of groups. This problem arise when I was studying Serre's "Trees".]

Comment by joseph

2012-01-11T04:26:59Z

As you have pointed out, for splitting fields, the theorem is true; no doubt about it. But I am not sure, whether it holds for non-splitting field. Is the answer "Exactly YES"?

Comment by joseph

2012-01-11T04:23:53Z

The isomorphism of vector spaces $Hom_H(V,V)\rightarrow Hom_G(Ind^G_H V, Ind^G_H V)$ will hold if $V$ is irreducible and $Res^H_{H\cap H_s} V$ and $Res^{H_s}_{H\cap H_s}V_s$ ($s\in G\setminus H$) do not share an irreducible component; what about converse?

Comment by joseph

2011-10-12T08:24:17Z

Thanks! Nice Answer!!

Comment by joseph

2011-10-12T04:08:50Z

@Torsten: Can you explain second paragraph in answer, please? (I know that Q8 has four 1-dim. irrep, and one 4-dim. irrep over Q; and over C (or Q(i)), it splits as sum of two irrep. of dim. 2 (i.e. twice the irrep. of dim 2 over $\mathbb{C}$))

Comment by joseph

2011-10-11T13:38:30Z

I am considering representations only on splitting fields, and looking whether there is smallest subfield (whether exist) in the algebraic closure $\bar{F}$. Not fixing any representation, since every irreducible representation over a splitting field remains irreducible even after extension of that splitting field, so not considering only one representation, but all irreducible representations. Is this the thing you want to know?

Comment by joseph

2011-09-10T07:44:55Z

***@Richard and Geoff***: Why should we consider "irreducible representations" of groups? In language of representation theory, the question will be simply "To find faithfull representations of finite groups over $\mathbb{R}$"; not necessarily irreducible.

Comment by joseph

2011-09-09T13:45:41Z

Given a subgroup $L$ of $G$, consider the set $X=\{H\leq G \colon H\cong L\}$. Does the cardinality of this set is $0$ (mod $p$) or $1$ (mod $p$)? For groups of order $p^2$, and $p^3$, it looks to be true.

Comment by joseph

2011-09-05T10:28:16Z

Also the Bass-Serre tree is just a tree; why we do not construct universal cover of graph of groups as again certain graph of groups with obvious universal property?

Comment by joseph

2011-09-05T10:28:12Z

@HW: The universal cover (Bass-Serre tree) of graph of groups is a tree on which the fundamental group of graph of group acts, with quotient graph of groups isomorphic to given graph of groups. But can we define it as a cover of given graph of groups which is also cover of any other cover of given graph of groups (similar to "topological universal cover")? I want to get all covers of graph of groups $\circ --\circ$, with vertex groups finite cyclic, edge group trivial.

Comment by joseph

2011-09-05T07:02:18Z

@David: Thanks! I was adding tag of "geometric group theory", but didn't get it. so added just group theory. Anyway!.

Comment by joseph

2011-09-05T06:51:37Z

As per the construction of Serre, the universal cover of a graph of groups is (just) a graph; so are we not moving away from category of graph of groups to category of graphs? Why we do not construct a universal cover as a graph of groups again?

Comment by joseph

2011-08-29T08:03:40Z

@HW: True! To avoid countably branching in $S$ or $\tilde{S}$, we can consider $S$ to be locally finite. Then can we hope for positive answer?

Comment by joseph

2011-08-29T07:09:25Z

@HW: Thanks! I have changed it.