# exercise in Isaacs's book on Character Theory

I'm stuck on an exercise in Isaacs's book "Character Theory of Finite Groups" - it relates to something I'm looking at as part of ongoing research, but I guess it belongs here rather than on MathOverflow, since it's an exercise and hence ought to be solvable by basic techniques...

Anyway, the question/problem is Exercise 5.14(a), and goes as follows.

Let G be a finite nonabelian group and let $f$ be the smallest character degree that is not 1. Suppose that the derived subgroup (a.k.a. the commutator subgroup) has order $\leq f$. Prove that the derived subgroup is contained in the centre of $G$.

The hint given is the (fairly obvious) fact that each conjugacy class injects into $G'$, so by the assumption of the problem is bounded above by $f$. This makes me think that the solution has something to do with column orthogonality in the character table, using the fact that the linear characters all take the value 1 on the derived subgroup; but I haven't managed to make that idea work.

If it helps or makes any difference: the chapter for which this is an exercise is the one introducing induced characters (but precedes the discussion of induction from normal subgroups, and Clifford theory). Again, I can't see a way to induce anything from the derived subgroup to make progress: my only idea would be to take a non-trivial irreducible character on the derived subgroup, induce it to a character on $G$, and observe that the induced character is orthogonal to all the linear characters of $G$ by e.g. Frobenius reciprocity.

I'm sure I'm just missing something obvious, so would be happy with just a hint or two.

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Welcome to math.SE, Yemon! But... where's the funky ice bear? –  t.b. Sep 18 '11 at 23:58
He is incognito –  Will Jagy Sep 19 '11 at 1:38
Incidentally, I think this question falls into the range where it's a good question on either math.SE or on MO. I didn't see it here, and heard about it on meta.MO. –  Noah Snyder Sep 25 '11 at 17:12

By the hint, as explained in Jack Schmidt's answer, you see that every conjugacy class has size smaller than f. Consider the conjugation action of G on the conjugacy class $g^G$. This is the sum of the trivial with a representation of dimension smaller than f. Thus, by assumption it is a sum of 1-dimensional representations, and therefore abelian. This means that $x y g y^{-1} x^{-1} = y x g x^{-1} y^{-1}$. Rearranging we get, $x^{-1} y^{-1} x y g = g x^{-1} y^{-1} x y$, and thus every commutator is central.

The reason this has to do with induced representations is that the conjugation action on $g^G$ is the representation induced from the trivial rep of the centralizer of g.

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The thought process here was that the best way to use the condition of the theorem is to use 'small dimension implies abelian.' Then it was just a matter of figuring out what small rep to look at, and that was already suggested by the hint and the end of Jack's answer. –  Noah Snyder Sep 25 '11 at 17:15
Very nice! Just a small correction: the dimension of the permutation representation is less than or equal to $f$. But it has a trivial summand, so the rest has dimension strictly smaller, and your argument goes through. –  Alex B. Sep 25 '11 at 23:32
Good point, I'll fix that. –  Noah Snyder Sep 26 '11 at 2:16
Thanks Noah! Given that the preceding problem in that section also concerned the conjugation representation, your solution is almost certainly the intended one. –  user16299 Sep 26 '11 at 20:00
If I've understood the argument correctly: it actually shows that when the largest conjugacy class size is $\leq f$, then all commutators are central - is that correct? –  user16299 Sep 26 '11 at 20:03
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Since no one has answered, I'll write down what I've had time to do:

The hint is proved as you say, but the injection is not just the identity, so I thought I'd mention it: If x and y are conjugate in G, then y = xg for some g in G, and so we get an element x−1y = [x, g] in G′. For a fixed x, we can recover y from c = x−1y as y = xc, and so the map from xG to G′ given by y maps to x−1y is injective, and |xG| ≤ |G′|. One does not generally get equality, since G′ consists of more than just commutators.

A few stronger claims are not true: S3 × S3 has a normal subgroup A3 × 1 whose size is less than the degree of a centerless character. The extra-special groups of order 32 have minimal non-linear character degree of 4, but have many non-central normal subgroups of order 4 (their derived subgroup and center have order 2). Thus it is important that the normal subgroup is contained in the derived subgroup.

The following seems to lead nowhere:

If |G′| ≤ f, then each conjugacy class size is less than f as well.

We want to show that if c in G′ and χ in Irr(G), then |χ(c)| = χ(1) ≥ f.

Column orthogonality gives: $$0 = \sum_{\chi \in Irr(G)} \chi(c)\chi(1) = [G:G'] + \sum_{\chi \in Irr(G) - Irr(G/G')}\chi(c)\chi(1)$$ Presumably now we take absolute values to finish, except I don't see anything useful. From column orthogonality, we also have $$\frac{|G|}{f} \leq |C_G(g)| = \sum_{\chi} |\chi(g)|^2 = [G:G'] + \sum_{\chi \in Irr(G)-Irr(G/G')} |\chi(g)|^2$$

A posteriori, we know that for every irreducible χ and g in G, |χ(g)| in { 0, χ(1) }, so we should get a lot of vanishing, but all of the inequalities I derived pointed the wrong way.

Inducing from the derived subgroup seems like a bad idea, since it is so small. Inducing from a centralizer might be reasonable, since its index is small, but I didn't see anything that actually helped.

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Thanks for the partial answer. I was thinking along the same lines you did, and wrote down the two equations/inequalities that you have; but I agree that it doesn't seem obvious how to finish offf –  user16299 Sep 22 '11 at 20:44
Yeah, my main reason for posting was to make it easier for someone to get started on it. The vanishing stuff is page 28, but I didn't see how to use it here. –  Jack Schmidt Sep 22 '11 at 20:52