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.
 A: 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.
A: 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.
