Character theory - Exercise 5.14

You should try and use the hint given!

Let $g \in G$. Then$|{\rm Cl}(g)|=|G:C_G(g)| \le |G'|$. Let $\chi$ be the trivial character of $C_G(g)$. Then the induced character $\chi^G$ has degree $|G:C_G(g)| \le |G'| \le f$.

Since the trivial character of $G$ is a constituent of $\chi^G$, it cannot be irreducible of degree $f$, so all of its constituents have degree less than $f$, and so they all have degree $1$. So $G' \le \ker( \chi^G)$ and, since $\ker (\chi^G) \le C_G(g)$, $G' \le C_G(g)$.

This is true for all $g \in G$, so $G' \le Z(G)$.

For (b) the statement should be: if $|[G,G']| \leq f$, then $G'$ is abelian. Note that the subgroup $[G,G']$ is also denoted as $\gamma_3(G)$, the third term of the lower central series of $G$ and is relevant in the study of nilpotent groups. It is the (normal, even characteristic) subgroup generated by the elements and its commutators: $\langle[g,x]: g \in G, x \in G' \rangle$. Maybe needless to say, but observe that $[G,G']=[G',G]$.

The proof of the statement above is a refinement of the proof that was put forward by Derek Holt.
Let $x \in \color{red}{G'}$ and consider $Cl_G(x)$, the conjugacy class of $x$. We can define a set-theoretic map $f: Cl_G(x) \rightarrow[G',G]$ by $$f(g^{-1}xg)=[x,g] (=x^{-1}g^{-1}xg).$$This map is easily seen to be injective. And it follows that for every $x \in G'$, we have $|Cl_G(x)|=[G:C_G(x)] \leq |[G,G']| \leq f$. Again as the former proof it follows that the induced trivial character $1_{C_G(x)}^G$ can only have linear constituents, whence $G' \subseteq ker(1_{C_G(x)}^G)=core_G(C_G(x)) \subseteq C_G(x)$, for all $x \in G'$. So $G' \subseteq \cap_{x \in G'}C_G(x)=C_G(G')$. This can only be the case if and only if $G'$ is abelian.

The argument can be generalized to show that if $i \geq 2$, and $|\gamma_{i+1}(G)| \leq f$, then $\gamma_i(G)$ is abelian (even $\gamma_i(G) \subseteq Z(G'))$.