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I was reading through a proof this morning which said that the characters of a group with abelian subgroup of index 2 are of degree at most 2. This feels like an easy result but I can't seem to work it out. I only have a basic understanding in group rep. theory so I looked through a few books on the subject - alas, no joy.

Could someone please help enlighten me?

Thanks

2 Answers 2

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Let us prove this in a series of steps. Firstly, we obviously need to understand the representations of abelian groups. The following exercise is important for this understanding:

Exercise 1 Prove that if $G$ is an abelian group, then every irreducible representation of $G$ is $1$-dimensional. (Hint: use Schur's lemma on the $G$-intertwining maps from a (complex) irreducible representation of $G$ to itself.)

Let us now consider a group $G$ with an abelian subgroup $H$ of index $2$. If $(\pi,V)$ is an irreducible representation of $G$, then $\pi$ induces a representation of $H$ by restriction which we denote by $(\pi_H,V)$. We consider two cases:

(1) If $(\pi_H,V)$ is irreducible, then $V$ is $1$-dimensional and the proof is complete.

(2) If $(\pi_H,V)$ is not irreducible, then there exists a minimal $H$-invariant subspace $W\subseteq V$. The following exercise allows us to understand $V$ in terms of $W$:

Exercise 2 If $g\in G$ and $g\not\in H$, then prove that $V=W+gW$. (Hint: remember that $(\pi,V)$ is an irreducible representation of $G$!)

Finally, we can determine a bound on the dimension of $V$:

Exercise 3 Prove that the dimension of $V$ is at most $2$. (Hint: the minimality of $W\subseteq V$ as a $H$-invariant subspace implies that $(\pi_H,W)$ is an irreducible representation of $H$; use Exercise 1.)

You could argue that the proof above is somewhat complicated. In fact, one can rewrite the above proof in a more conceptual manner which allows one to see the real idea of the proof. The point is essentially the relationship between the representations of a group and the induced representations of subgroups and, where applicable, quotient groups.

I hope this helps!

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    @user32259 In fact, for more information on accepting an answer, you may wish to have a look at http://meta.math.stackexchange.com/questions/3286/how-do-i-accept-an-answer.2012-05-26
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One of the most important results in representation theory is Frobenius reciprocity: if $H\leq G$ are finite groups, $\phi$ is a character of $G$ and $\psi$ is a character of $H$, then $\langle\rm{Ind}_{G/H}\psi,\phi\rangle = \langle\psi,\rm{Res}_{G/H}\phi\rangle$, where $\langle-,-\rangle$ denotes inner product of characters. It immediately follows that any irreducible character of $G$ is a summand of some character induced from $H$. Recall that under induction, the degree of a character increases by a factor of $[G:H]$. Finally, using that all irreducible characters of abelian groups are 1-dimensional, you obtain

Proposition: If $G$ has an abelian subgroup of index $n$, then all irreducible characters of $G$ are of degree at most $n$.

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    thanks, this is interesting. i haven't covered induced and restricted representations yet but thank you for putting the result in a more general context.2012-05-26