# Finite groups with lots of conjugacy classes, but only small abelian normal subgroups?

Denote the commuting probability (the probability that two randomly chosen elements commute) of a finite group $$G$$ by $$\operatorname{cp}(G)$$. By a result of Gustafson , $$\operatorname{cp}(G)=\operatorname{k}(G)/|G|$$, where $$\operatorname{k}(G)$$ is the number of conjugacy classes of $$G$$. My question is the following:

Does there exist a function $$f:\left(0,1\right]\rightarrow\left[0,\infty\right)$$ such that the following holds: "If $$G$$ is a finite group, $$\rho\in\left(0,1\right]$$, and $$\operatorname{cp}(G)\geq\rho$$, then $$G$$ has an abelian normal subgroup of index at most $$f(\rho)$$."?
Equivalently: Does there exist no sequence $$(G_n)_{n\geq 1}$$ of finite groups such that $$\operatorname{cp}(G_n)$$ is bounded away from $$0$$, yet the minimal index of an abelian normal subgroup in $$G_n$$ converges to $$\infty$$ as $$n\to\infty$$?

This question arose while I was thinking about different definitions of "almost-abelianity" of finite groups $$G$$, each of which generalizes abelianity in a quantitative way and which I briefly discuss now for the sake of motivation.

1. For $$I,C\in\left[0,\infty\right)$$, say that $$G$$ is $$(I,C)$$-almost abelian if and only if $$[G:\operatorname{Fit}(G)]\leq I$$ and $$\operatorname{cl}(\operatorname{Fit}(G))\leq C$$. Here, $$\operatorname{Fit}(G)$$ denotes the Fitting-subgroup of $$G$$, and $$\operatorname{cl}(H)$$ the nilpotency class of a nilpotent group $$H$$.
2. For $$\rho\in\left(0,1\right]$$, say that $$G$$ is $$\rho$$-almost abelian if and only if $$\operatorname{cp}(G)\geq\rho$$.
3. For $$I\in\left[0,\infty\right)$$, say that $$G$$ is $$I$$-virtually abelian if and only if $$G$$ has an abelian normal subgroup of index at most $$I$$.

Each of the following facts about finite groups $$G$$ is either not difficult to see or follows from results in the literature:

1. Virtual abelianity implies almost-abelianity in the second sense." More precisely, there is a function $$f:\left[0,\infty\right)\rightarrow\left[0,1\right)$$ such that if $$G$$ is $$I$$-virtually abelian, then $$G$$ is $$f(I)$$-almost abelian. Actually, one can choose $$f:I\mapsto I^{-2}$$.
2. "Almost-abelianity in the second sense implies almost-abelianity in the first sense." More precisely, there are functions $$f_1,f_2:\left(0,1\right]\rightarrow\left(0,\infty\right)$$ such that if $$G$$ is $$\rho$$-almost abelian, then $$G$$ is $$(f_1(\rho),f_2(\rho))$$-almost abelian (the existence of $$f_1$$ is by [1, Theorems 4(iii) and 8(ii)], and the existence of $$f_2$$ follows from [1, Lemma 2(ii)] and the fact that $$\operatorname{cp}(G)\leq 5/8<1$$ for all nonabelian finite groups $$G$$ ).
3. "Almost-abelianity in the first sense does not imply almost-abelianity in the second sense." More precisely, even if both the index and the class of the Fitting subgroup of $$G$$ are bounded, $$\operatorname{cp}(G)$$ can be arbitrarily small.

Hence the later a property is defined in the list above, the stronger it is, and the second property is strictly stronger than the first. My question asks whether the third property is also strictly stronger than the second.

References:

 W. H. Gustafson, What is the probability that two group elements commute?, Amer. Math. Monthly 80 (1973), 1031-1034.
 R. M. Guralnick and G. R. Robinson, On the commuting probability in finite groups, J. Algebra 300 (2006), 509-528.

• YCor answered your question, but I note that YCor's example has very few commutators. Since "probability that two elements commute" is "probability that a random commutator is the identity," it might be reasonable to ask what happens if you take "almost-abelian" to mean "bounded number of commutators."
– JSE
May 9, 2016 at 20:06
• @JSE: This also looks like an interesting condition. I will think about it, thank you! May 9, 2016 at 20:56
• @YCor has already answered your question, but you might be interested in this old question I asked mathoverflow.net/questions/21071/… and the answer given by "BugsBunny". The question concerned a "fourth" version of being close to abelian, namely that all irrreducible characters have small degree; I think this is equivalent to your 3., but the bounds aren't very good May 9, 2016 at 21:06
• @JSE no, there are trivial examples with an unbounded commutator group, for instance $C_3^n\rtimes C_2$ where $C_k$ is cyclic of order $k$, $C_2$ acts on $C_3^n$ by $x\mapsto -x$ (all elements of $C_3^n$ are commutators). Of course it has an abelian subgroup of bounded order, but you can fix this by taking the direct product with the group $G_n(q)$ of my example and both the derived subgroup and the smallest index of an abelian subgroup tend to infinity.
– YCor
May 10, 2016 at 6:18

No: there exists a sequence of finite groups with commuting probability bounded away from 0 but with no abelian (normal) subgroup of bounded index.

Fix a prime power $q$. Consider the "higher Heisenberg" group $G_n$ of order $q^{2n+1}$ consisting of those square matrices of size $n+2$ over $\mathbf{F}_q$ of the form $$P(u,v,x)=\begin{pmatrix}1 & u & x\\ 0 & I_n & v\\ 0 & 0 & 1\end{pmatrix},$$ where $u$ is a row, $v$ is a column, and $x$ a scalar. If we endow $\mathbf{F}_q^{n^2}$ with the symplectic product $\langle u\oplus v,u'\oplus v'\rangle=uv'-u'v$, then we see that the centralizer of $P(u,v,x)$ is the set of $P(u',v',x')$ such that $\langle u\oplus v,u'\oplus v'\rangle=0$. In particular, this is a subgroup of index $q$ (unless $(u,v)=(0,0)$ in which case $P(u,v,x)$ is central). So the probability that two elements commute is $\ge 1/q$ [actually it's $1/q+q^{-2n}(1-1/q)$, if I'm correct].

On the other hand, the largest cardinal of an abelian subgroup in $G_n$ is $q^{n+1}$ (since $n$ is the largest dimension and hence $q^n$ is the largest cardinal of an isotropic subspace in the symplectic space $\mathbf{F}_q^{2n}$). So the minimal index of an abelian subgroup in $G_n$ is $q^n$. Thus, $q$ being fixed ($q=2$ is fine) and $n$ tending to infinity, we get the required example.

• Are these examples of extraspecial $q$-groups? (I had a feeling that they would give a counterexample, but couldn't quite get the right bounds on the index of an abelian subgroup) May 9, 2016 at 21:04
• @YemonChoi: it's extraspecial iff $q$ is prime
– YCor
May 10, 2016 at 6:53
• These examples have commutator subgroup of bounded size. However, say assuming $q$ is odd, if we allow the 1 in the upper left and lower right to be replaced by $\pm 1$ (so we get a group $\{\pm 1\}^2\ltimes G_n$), the commuting probability is $\ge 1/16q$ and the derived subgroup is the subgroup $G_n$ of index 4, and abelian subgroups have index $\ge q^n$ (probably $\ge 4q^n$).
– YCor
May 10, 2016 at 6:58

Let $G$ be an extraspecial group of order $p^{2n+1}$. Then $$cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}.$$ All maximal abelian subgroups have index $p^n$ in $G$. Hence, the answer is `not'.