Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 3 down vote accepted

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!

share|cite|improve this answer
Many thanks Amitesh. I found the steps very useful. Your help is much appreciated! P.S. I'm new to this site - is there a more fomral way of thanking you like clicking a link somewhere? – user32259 May 26 '12 at 13:08
@user32259 You are welcome! If you are happy with my answer above, then you can "accept my answer" by clicking the tick mark just below the vote count (the thing with an up arrow, a down arrow, and a number in the middle to the left of my answer; there is a tick mark just below the down arrow). In general, you can also upvote an answer by clicking the arrow pointing upward to the left of my answer (accepting an answer and upvoting an answer are independent of each other; you can upvote any answer you like but you can accept only one). – Amitesh Datta May 26 '12 at 13:42
@user32259 In fact, for more information on accepting an answer, you may wish to have a look at…. – Amitesh Datta May 26 '12 at 13:43

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 $n$.

share|cite|improve this answer
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. – user32259 May 26 '12 at 13:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.