Relation between the index of a group and its center and the order of the commutator

Question

I've been studying representation theory and a doubt has originated within me

I'm working with nonlinear irreducible characters, and by exercise 2.10 on Isaac's "Character theory of finite groups" I know that if $\chi (1) > 1$ then $\chi(1) \geq |G|/(n-1)$ with $n = |G:A|$ with $A$ an abelian subgroup of $G$.

But we know that the center of a group $Z(G) \subseteq G$ is abelian, by its definition, so $|G:Z(G)|=n$

Now, what I'd like to know is the relation of this index with the order of the conmutator subgroup $G'$ defined in the book as \begin{align} G' = \cap \{Ker(\chi) | \chi \in Irr(G), \chi(1)=1\} \end{align}

\begin{align} |G:Z(G)| ??|G'| \end{align}

If there's a relation from group theory it would also be helpful, it doesn't have to come from the definition I gave previously

Pd: the reason why I cited exercise 2.10 is that the relation involving the $n-1$ would be useful in the problem I'm working on

Thank you in advance, I hope there exists such relation.

Edit: I forgot to mention that I have that $|G'|=p$ with $p$ a prime number.

Answer 1

There is no clear relation between $[G:Z(G)]$ and $|G'|$. For the Dihedral group $D_{2n}$, of order $2n$, $n$ odd, the former number is $2n$ and the latter is $n$. For Abelian groups and simple groups the numbers are equal. If $p,q$ are odd primes, $p\mid q-1$, then there is a non-abelian group of order $pq$ with the first number equal to $pq$ and the second number equal to $q$. For the Heisenberg group of order $p^3$, the first number is $p^2$ and the second number is $p$.

Edit. Influenced by Derek Holt's answer, I found a connection between $|G/Z(G)|$ and $|G'|$: If $|G/Z(G)|=n$ then $|G'|\le n^{2n^3}$.

Answer 2

One imprecise but useful result connecting these quantities is that $g^{|G:Z(G)|}=1$ for all $g \in G' \cap Z(G)$, so the exponent of $G' \cap Z(G)$ divides $|G:Z(G)|$.

You get this by considering the transfer homomorphism $G \to Z(G)$, which has $G'$ in its kernel and maps elements of $Z(G)$ to $g^{|G:Z(G)|}$.

One application of this is a result of Schur which says that if $G$ is a (possibly infinite) group with $|G:Z(G)|$ finite, then $G'$ is finite.

In fact, if we let $n=|G:Z(G)|$, then $G'$ is generated by at most $n^2$ elements (we can take the set of all $[g,h]$ with $g,h$ coset reps of $Z(G)$ in $G$) and, since $|G':G' \cap Z(G)| \le n$, $G' \cap Z(G)$ is generated by at most $n^3$ elements, and hence $|G' \cap Z(G)| \le n^{n^3}$ and $|G'| \le n^{n^3+1}$, which is better than the bound mentioned by markvs. But perhaps I have made a mistake!

Answer 3

In addition to what Derek Holt wrote, if $G$ is finite and $G/Z(G)$ is a $p$-group, then $G'$ is a $p$-group. But for me it is unclear what exactly are you asking. Are you looking for a solution to Problem (2.10) in Isaacs' book (CTFG)? You assume that $|G'|=p$. Well, if this is the smallest prime dividing $|G|$ then

$(a)$ $ p \mid |G:Z(G)|$ and $p \mid |Z(G)|$
$ (b) $ $|G:Z(G)|$ is a square.

Proof of $(a)$ Since $G/C_G(G')$ embeds in $Aut(C_p) \cong C_{p-1}$ and $p$ is the smallest prime with $p \mid |G|$, it follows that $G' \subseteq Z(G)$ and hence $p \mid |Z(G)|$. Of course $G$ is not abelian ($G' \cong C_p$), so one can find a $g \in G-Z(G)$. Then $1 \lt \#Cl_G(g) \leq |G'|=p$ and $\#Cl_G(g)=|G:C_G(g)| \mid |G|$, so since $p$ is the smallest prime dividing $|G|$, it follows that $\#Cl_G(g)=p$. But, $Z(G) \subsetneq C_G(g) \subsetneq G$, so $p \mid |G:Z(G)|$ (and even $|G:Z(G)| \geq p^2$).
The proof of $(b)$ requires some character theory, and I skip it here, since probably I am drifting off (I can add it later if you wish).