# Finding the cohomology class of an irreducible representation of a normal subgroup

Dear Mathoverflow,

I have a question concerning finite dimensional irrducible representations of normal subgroups and how to construct representations of the supergroup out of these representations with a computer (using GAP). The following text describes a construction, the question itself is on the bottom. Let $G$ be a finite group and let $N \triangleleft G$ be a normal subgroup (which is necessarily a finite index subgroup). Now let $\vartheta: N\to \mathbb{C}$ be an irreducible character. My goal is to construct all irreducible characters $\pi$ of $G$ whose restriction to $N$ is equal to a multiple of $\vartheta$. My references for this is the article "Representations Induced in an Invariant Subgroup" written by Alfred Clifford in 1937 and the book "Character theory of finite groups" by Martin Isaacs.

The book of Isaacs (chapter 11) gives an explicit way for this construction: First we find the inertia subgroup of $\vartheta$ relative to $G$, that is the subgroup of $G$ defined by $\langle g\in G| \vartheta(h^g) = \vartheta(h)\ \forall h\in N\rangle$. Then we take an irreducible representation $\rho$ of $N$ affording the character $\vartheta$ and choose a projective representation $P$ such that the restriction of $P$ to $N$ is equal to $\rho$. This projective representation can be contructed for example by taking a coset transversal $T$ of $N$ in $G$ and then take for every $t\in T$ a matrix $P_t\in GL_n(\mathbb{C})$ such that

$P_t^{-1}\rho(h)P_t = \rho(t^{-1}ht)\ \forall h\in N$

and define the projective representation $P$ to be $P(th) = P_t\rho(h)$. Since $P$ is a projective representation, there is a well-defined 2-cocycle $\alpha\in Z^2(G/N, \mathbb{C}^* )$ defined via $\alpha(gN,hN) := P(g)P(h)P(gh)^{-1}$. This 2-cocycle has a cohomology class in the Schur-Multiplier

$H^2(G/N, \mathbb{C}^* ) = Z^2(G/N, \mathbb{C}^* )/ B^2(G/N, \mathbb{C}^* )$

and we take a projective representation $\beta$ of $G/N$ with a cocycle cohomologous to the inverse to $\alpha$, take the tensor product $P\otimes \beta$ and then take the induced representation of $P\otimes \beta$ to get an irreducible representation of $G$.

Each of the steps above is realisable by a computer, I'm pretty sure. There is just one step I don't know how to do it. If we have a cocycle $\alpha$ which is a function from $(G/N)^2\to \mathbb{C}^*$ then we want to know which cohomology class it actually has in $H^2(G/N, \mathbb{C}^* )$. Since $H^2(G/N, \mathbb{C}^* )$ is abelian and finite in this case, there is an isomorphism between $H^2(G/N, \mathbb{C}^* )$ and its irreducible characters. So we can assign an irreducible character of $H^2(G/N, \mathbb{C}^* )$ to every 2-cocycle in $Z^2(G/N, \mathbb{C}^* )$.

My question is: How can I assign the irreducible character to a 2-cocycle? I know that this character exists and I know how to construct the 2-cocycle, but I don't know how to assign a character to this 2-cocycle with an algorithm.

Another question might be: Is there an easier way to find the cohomology class corresponding to $\vartheta$?

A third way would be to have the 2-cocycle $\alpha$ and for every cohomlogy class we have a representative $\gamma$. Then we can check whether $\alpha\gamma^{-1} \in B^2(G/N,\mathbb{C}^* )$ or not. There I also don't know how to do this.

Maybe someone here can help me. Thank You very much!

-
Magma can handle 2-cocycles and 2-coboundaries, see here: magma.maths.usyd.edu.au/magma/handbook/text/703 . – Matthias Künzer Sep 22 '11 at 17:52
Perhaps one will have more chances with this question in the GAP mailing list. – Pasha Zusmanovich Sep 23 '11 at 6:33
Probably the easiest way to use a computer to find all of the irreducible characters $\chi$ of $G$ whose restriction to $N$ has the given character $\theta$ as an irreducible constituent is simply to ask the software to compute the induced character $\theta^G$. The irreducible constituents of that induced character are exactly the characters you are looking for. – Marty Isaacs Jan 28 '12 at 21:28