Character theory - Exercise 5.14

Question

I am trying to solve the exercise 5.14 from the Isaac Martins Character Theory of Finite Groups.

Let $G$ be a nonabelian group and let $ f={\rm min}\{\chi(1) \,\,|\,\, \chi \in {\rm Irr}(G), \chi(1)>1 \}. $ Show that:

(a) If $ |G'| \leq f,$ then $G'\subset Z(G)$
(b) If $[G:G'] \leq f,$ then $G'$ is abelian
(c) If $H\subset G$, and $[G:H]\leq f$, then $G' \subset H$

For (a), the indication given is: if $x\in G$, then $|Cl(x)|\leq |G'|$

What I have tried for (a) is the second orthogonality relation $$|G|=\sum_{\chi_i}|\chi_i^2(g)|=[G:G']+\sum_{\chi_i {\rm nonlinear}}|\chi_i^2(g)| .$$ For $g=1$, it gives $$ |G|=[G:G']+\sum_{\chi_i {\rm nonlinear}}|\chi_i^2(1)|\geq [G:G'] +\sum_{\chi_i {\rm nonlinear}}|f|^2=[G:G']+(k-[G:G'])|f|^2$$
where $k$ is the number of conjugacy classes. Nothing much from that.

What I know as well is $|G|\geq |Z(G)|\chi(1)^2$ for any $\chi \in {\rm Irr}(G)$ (derived from the first orthogonality relation with $g \in Z(G) \Rightarrow |\chi(g)|=\chi(1)$). Also $\chi(1) | [G:Z(G)]. $

I tried to show $G/Z(G)$ abelian but no success either. Can someone please help me ?

Answer 1

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 $\gamma_i(G) \subseteq Z(G'))$.

Answer 2

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 $G' \le Z(G)$.