Alternative title: “an analyst does some finite group theory, latest in an irregular and infrequent series”. Or: when life gives you lemons, get your engineers to invent a combustible lemon …

Shortly after posting the preprint mentioned in this blogpost to the arXiv, at the tail end of 2020, I started to find minor errors or omissions in the exposition; none of them affect the validity of the main results, but it did mean that some major revisions were needed. Unfortunately, the demands of teaching and assessment and quality assurance (the joys of UKHE) left me with little time and even less energy to focus on doing so.

Instead, since I was teaching a first course on groups (and rings), I found myself playing around with character theory of finite groups to keep myself entertained, and it was during one of these spells when I realized how a result in that paper could be sharpened for finite groups. Subsequently, I worked out a way to obtain the sharper result for all virtually abelian groups, but the proof becomes more technical and has more moving parts, because one does not have access to the same counting arguments that are possible for finite groups.

The proof for the general case is being written up, and will be added to the revised version of the arXiv preprint. But it seemed worth writing up the simpler argument for the finite case, for two reasons: increased forgetfulness as I get older; and increased sentimental attachment as I get older. So here it is. Probably it should have been split over several posts, but then that would have greatly decreased the chances of ever finishing it.

For a finite group G, let Irr(G) be the set of irreducible characters of G (throughout, we are working over the complex numbers, so no modular representation theory to see here). If $\phi\in {\rm Irr}(G)$ then we write $d_\phi=\phi(e)$ for the degree of φ, or equivalently, the dimension of any (irreducible) representation of which φ is the trace. A fundamental result in the classical character theory of finite groups is that $|G| = \sum_{\phi\in {\rm Irr}(G)} (d_\phi)^2$.

Consider the following quantity, which emerged in work of Johnson (J. London Math. Soc. 1994) on the (non-)amenability of Fourier algebras of compact groups, but which in the case of finite groups can be defined directly:

$\displaystyle{\rm AD}(G) := |G|^{-1} \sum_{\phi\in {\rm Irr}(G)} (d_\phi)^3$

Remark 1. The notation AD stands for “antidiagonal”, and does not appear in Johnson’s original paper. The reason for this notation/terminology comes from the 2020 work mentioned above, where AD(G) is defined in a more abstract way — but to keep this blog post more focused, I will not go into this here.

It is my belief that this numerical invariant of G deserves further study. It is fairly obvious from the definition above that ${\rm AD}(G\times H)={\rm AD}(G)\, {\rm AD}(H)$. What is much less obvious, to me at least, is that whenever H is a subgroup of G, we have AD(H) ≤ AD(G). Indeed, I do not know of a direct proof using the definition above, given that the representation theory of a subgroup of G can be radically different from that of G itself; the only proof I know goes via the “abstract definition” that is alluded to above.

So what can we say about AD(G)? Let us start with some results that can be found in Johnson’s 1994 paper. First of all, since $d_\phi \geq 1$ for all φ,

$\displaystyle {\rm AD}(G) = \frac{1}{|G|} \sum_{\phi\in {\rm Irr}(G)} (d_\phi)^3 \geq \frac{1}{|G|} \sum_{\phi\in {\rm Irr}(G)} (d_\phi)^2 = 1$

and this inequality is strict unless $d_\phi=1$ for all $\phi\in {\rm Irr}(G)$. That is:

Proposition 2.
For any finite group $G$, we have ${\rm AD}(G)\geq 1$. Equality holds if and only if $G$ is abelian.

This raises a natural question: if G is a non-abelian finite group, how small can $AD(G)$ be?

On the face of it, is conceivable that one could find non-abelian finite groups $G_n$ such that $AD(G_n) \searrow 1$. But inspecting the argument used to prove Proposition 2, one sees that for $AD(G)$ to be close to 1, there must be a high proportion of 1-dimensional characters among the elments of Irr(G). Now every 1-dimensional character factors through the abelianization $G\to G_{ab}$; more precisely, the 1-dimensional characters of G correspond to the group homomorphisms $G_{ab}\to {\mathbb T}$, and hence by Fourier analysis for finite abelian groups, there are exactly $|G_{ab}|$ of these characters. Since $G_{ab}$ is (isomorphic to) the quotient of G by its commutator subgroup [G.G], and since [G,G] has size at least 2 when G is non-abelian, we see that

$\displaystyle | \{ \phi \in {\rm Irr}(G) \colon d_\phi= 1\} | = \frac{|G|}{|[G,G]|} \leq \frac{|G|}{2}$

and from here, it is a short step to the following result.

Theorem 3. If $G$ is a non-abelian finite group, then ${\rm AD}(G) \geq 3/2$.

Before explaining the proof of the theorem, we note the following corollary.

Corollary 4. If $H_1, \dots, H_n$ are finite non-abelian groups, then ${\rm AD}(H_1\times \dots \times H_n) \geq (3/2)^n$.

In particular, by taking powers of some fixed class-2 nilpotent group, we see that there are class-2 nilpotent groups with arbitrarily large AD-constant.

The proof of Theorem 3. Let $\Omega_n$ denote the set of irreducible characters of G which have degree n, so that

$\displaystyle 1 = \frac{1}{|G|} \sum_{n\geq 1} n^2 |\Omega_n| \hbox{ and } {\rm AD}(G) = \frac{1}{|G|} \sum_{n\geq 1} n^3 |\Omega_n|$

Thus

and rearranging gives ${\rm AD}(G) \geq 2 - |G|^{-1}|\Omega|$. But since $G$ is non-abelian, we know from the remarks before the theorem that $|G|^{-1}|\Omega_1|\leq 1/2$, and the result follows.

So far, all of this is in Johnson’s original paper. However, he stops short of characterizing those non-abelian G for which the extremal value AD(G)=3/2 is attained. Inspecting the proof of Theorem 3, one sees that in order for AD(G) to equal 3/2, the following conditions are both necessary and sufficient:

• Equality must hold in (*);
• $|G|^{-1} |\Omega_1| = 1/2$.

In turn, these are equivalent to the following pair of necessary and sufficient conditions:

Second set of conditions:

It turns out that these conditions can be replaced with one that is purely group-theoretic, i.e. one that does not make any reference to characters.

Theorem 5. Let G be a finite group. TFAE:

2. G/Z(G) is isomorphic to the Klein-four group.

The last condition may be phrased as: G is a non-trivial central extension of $C_2\times C_2$. It should therefore be possible to classify all possible G using the methods of group cohomology, but I haven’t made serious efforts to look up the necessary details.

Remark 6. Instinctively, one feels that $2)\implies 1)$ should be easier than $1\implies 2$, because 2) specifies some structural property of G and 1) is a statement about some numerical invariant of G defined in terms of its representation theory. However, during my original investigations in March/April 2021, it was the implication $1)\implies 2)$ which came first, because as we will see it follows quite easily if one takes for granted certain basic facts about the character table of a finite group. The converse implication is conceptually easy — since Z(G) has small index in G, the commutators in G must have a restricted form — but somehow writing out the proof involves more nitpicking over cases than I expected.

#### Proving the implication $1)\implies 2)$

We start from the “second set of conditions” stated above, and use them to show that $|G:Z(G)|=4$. If we can show this, then G/Z(G) is a group of order 4; and up to isomorphism the only groups of order 4 are $C_4$ and the Klein-four group. But now we have a general group-theoretic fact: if H is any group (not necessarily finite), N is a subgroup of Z(H), and H/N is cyclic, then H is abelian. (Students who I was teaching in the first half of 2021 may recognize this from the mock exam paper!) In particular, in our setting $G/Z(G)$ cannot be cyclic, so it has to be isomorphic to the Klein-four group.

Let us now show that $|G:Z(G)|=4$. The idea here is quite natural if one has ever spent time playing with character tables; note that by the given assumptions on G, all irreducible characters have degree 1 or 2, and we have a great deal of control on the number of degree 1 characters. To be precise, using the notation above, we know that $|{\rm Irr}(G)| = |\Omega_1| + |\Omega_2|$ and we also know that $|G| = |\Omega_1| + 4|\Omega_2|$. Since | [G,G] | =2, the order of $G_{ab}$ is half the order of G, and by Fourier/Pontryagin duality this means $|\Omega_1| = \frac{1}{2}|G|$. Substituting this into the previous equations we obtain

$\displaystyle |{\rm Irr}(G)| = |\Omega_1| + \frac{1}{4}( |G| - |\Omega_1|) = \frac{5}{8}|G| \;.$

On the other hand: let us try to count conjugacy classes in G. In any group H, there is an injection from each conjugacy class into the set of commutators of H, so in particular the size of every conjugacy class is bounded above by the order of the commutator subgroup. Since we are assuming that |[G,G]|=2, it follows that each conjugacy class of G is either a singleton (i.e. an element of the centre) or has size exactly 2. Writing $k_2$ for the number of conjugacy classes of size 2, we have $|G|= |Z(G)|+ 2k_2$ and

$\displaystyle |{\rm Conj}(G)| = |Z(G)|+k_2 = |Z(G)| + \frac{1}{2}(|G|-|Z(G)|) = \frac{1}{2}|G| +\frac{1}{2}|Z(G)| \;.$

But the character table is a square! That is, $|{\rm Irr}(G)|=|{\rm Conj}(G)|$ (one of the most notorious “unnatural bijections” in algebra). We therefore have

$\displaystyle \frac{5}{8}|G| = \frac{1}{2}|G| +\frac{1}{2}|Z(G)|$

and rearranging gives $|G|= 4 |Z(G)|$ as required.$\hfill \Box$

#### Proving the implication $2)\implies 1)$

It suffices to show that 2 implies both of the “second set of conditions“. Pick a non-central element $x_0\in G$, and let $H$ be the subgroup of G generated by $x_0$ and Z(G). Then H is contained in the centralizer $Z_G(x_0)$, which is a proper subgroup of G since $x_0$ is non-central, and so $|G:H|\geq |G:Z_G(x_0)| \geq 2$. On the other hand, Z(G) is a proper subgroup of H since $x_0\in H$, and so $|H:Z(G)|\geq 2$. Hence

$\displaystyle 4 = |G:Z(G)| = |G:H|\, |H:Z(G)| \geq 2 |H:Z(G)| \geq 4.$

Since equality must hold throughout, we see that $|G:H|=2$. Now $H$ is abelian, so all its irreducible characters have degree 1. If $\phi\in {\rm Irr}(G)$, then by considering ${\phi\vert}_H$ and invoking Frobenius reciprocity, we see that φ must occur as a summand of some induced character ${\rm Ind}^G_H \chi$ where $\chi \in {\rm Irr}(H)$. Since H is abelian $d_\chi=1$, and so $d_\phi \leq |G:H|d_\chi =2$. This establishes the first of the two conditions.

It remains to show that $| [ G,G] | =2$. What follows is not the original proof that I came up with — see Remark 7 below — but it preserves more symmetry, at the expense of being less direct.

Let $q:G\to G/Z(G)$ be the quotient map. The Klein-four group has the following properties:

• each non-identity element has order 2 (which by a favourite exercise of those teaching group theory, implies the group is commutative);
• any two distinct non-identity elements generate the whole group.

Now partition $G$ into four distinct cosets of $(G)$, labelled as $Z(G), aZ(G), bZ(G), cZ(G)$. For convenience, let $x\sim y$ denote the equivalence relation “x and y belong to the same coset of Z(G)” (equivalently, $q(x)=q(y)$). Then the properties of the Klein-four group listed above imply:

• $a\sim a^{-1} \;,\; b\sim b^{-1} \;,\; c\sim c^{1}$;
• $ab\sim c\sim ba \;,\; bc \sim a \sim cb \;,\; ca \sim b \sim ac \;.$

The equivalence relation respects multiplication (since q is a homomorphism) and so we also have $b\sim a^{-1}c$, etc.

The key point: a little thought (or exercise) shows that if $x\sim x'$ and $y\sim y'$ then $[x,y]=[x',y']$. (Here it is important that we are quotienting by Z(G) and not by some arbitrary normal subgroup.)

Since $a\sim a^{-1}$ and $b\sim ac$, we have

$\displaystyle [a,b] = [a^{-1},ac] = a^{-1}(ac) a (ac)^{-1} = [c,a]$;

and since $a\sim cb\sim cb^{-1}$, we also have

$\displaystyle [a,b] = [cb^{-1},b] = (cb^{-1})b(bc^{-1})b^{-1}=[c,b]$

Everything said thus far remains invariant under a permutation of the symbols $a,b,c$, since this just corresponds to a relabelling of the non-trivial cosets of Z(G). Therefore, starting from the identities $[a,b]=[c,a]=[c,b]$, we obtain

$\displaystyle [b,c] = [a,b]=[a,c] \quad\hbox{and}\quad [c,a] = [b,c]=[b,a].$

Thus all six expresions $[a,b], [b,c],[c,a],[b,a], [c,b], [a,c]$ are equal to the same element of G, which we denote by $z_0$. Since every element of G is equivalent to one of $\{e,a,b,c\}$, we have shown that $[x,y] \in \{e,z_0\}$ for all $x,y\in G$.

Note that $z_0\neq e$. For since $G\neq Z(G)$, there exist $x,y\in G$ with $xy\neq yx$. But $x$ is equivalent to exactly one element of the set $T:=\{a,b,c\}$ and $y$ is equivalent to another element of $T$, so $[x,y]=z_0$.

Moreover,

$\displaystyle [a,b] = [b,a] =[b^{-1},a^{-1}] = ([a,b])^{-1}$

which shows that $z_0$ is an involution. Thus $\{e,z_0\}\subseteq [G,G] \subseteq \langle z_0 \rangle = \{e,z_0\}$ and we have shown that $|[G,G]=2$ as required. $\hfill \Box$

Remark 7. The proof just given avoided “breaking symmetry”, in the sense that we did not privilege any of the non-trivial cosets of Z(G) over any of the others; equivalently, when considering G/Z(G) we did not pick two specific generators. However, it did rely on some ad hoc trickery using the equivalence relation $\sim$ to show that all non-trivial commutators take the same value. In this context, it should be admitted that the argument above is not the one I first came up with when first proving $2)\implies 1)$ in Theorem 5. The original argument went as follows: fix two noncommuting elements $a_0$ and $b_0$ in G, and note that their images in G/Z(G) commute and generate the whole group, so that G can be written as a disjoint union $Z(G) \cup a_0Z(G) \cup b_0Z(G) \cup a_0b_0Z(G)$ with $b_0a_0Z(G)=a_0b_0Z(G)$. Writing $z_0 = [a_0,b_0]$ one has $z_0b_0 = a_0 b_0 (a_0)^{-1}$; squaring both sides, and using centrality of $z_0$ and $b_0^2$, we obtain

$\displaystyle (z_0)^2(b_0)^2 = (z_0b_0)^2 = a_0 (b_0)^2 (a_0)^{-1}=(b_0)^2$

Thus $z_0$ is an involution. I then did a case by case analysis of the various commutators $[a_0,b_0], [b_0,a_0b_0]=[b_0,b_0a_0]$, etc. and used manipulations similar to those above to show that all non-trivial commutators were equal to $[a_0,b_0]$. Compared with the proof above, this approach feels slightly more “hacky”, but it does seem to suggest more naturally why one might hope to reduced all the commutators to (expressions involving) $[a_0,b_0]$.

22 September, 2021 8:44 pm

It is well known that, in a finite non-abelian group G, the probability that two elements commute is at most 5/8. Moreover, it is not hard to show that equality is achieved if and only if G/Z(G) is isomorphic to the Klein-four group. So this is another characterisation of the class of groups that you are interested in.

Perhaps there is a more general relation between AD(G) and the commuting probability? (I haven’t thought much about it.)

• 22 September, 2021 9:57 pm

Thanks! I might add your observation, suitably credited, to the “concluding remarks and further questions” section of the paper that I am revising — would that be OK?

There is at least one “cheap” connection between AD(G) and cp(G), as remarked upon in this MO question: Hölder’s inequality tells us that $1\leq {\rm cp}(G) {\rm AD}(G)^2$. Feeding in the condition ${\rm AD}(G)=3/2$ only gets us ${\rm cp}(G) \geq 4/9$, which is not enough to allow us to leverage the result concerning maximal cp.

As I remarked in comments on that MO question, cp plays well with taking quotients and AD plays well with taking subgroups. So there seems to be an informal dictionary of heuristics, but the inequality I gave above is too weak to allow translation of actual results. Of course, that inequality is far from sharp, so perhaps one could do better.