The n-Category Café

In fact everything depends on what field $k$ we’re using for our vector spaces! For starters let’s take $k = \mathbb{Q}$.

Here’s a list of finite groups where the map $\beta : A(G) \to R(G)$ from the Burnside ring to the representation ring is known to be surjective, taken from the nLab article Permutation representations:

1. cyclic groups,

2. symmetric groups,

3. $p$-groups (that is, groups whose order is a power of the prime $p$),

4. binary dihedral groups $2 D_{2n}$ for (at least) $2 n \leq 12$,

5. the binary tetrahedral group, binary octahedral group, and binary icosahedral group,

6. the general linear group $GL(2,\mathbb{F}_3)$.

Now, these may seem like rather special classes of groups, but devoted readers of this blog know that most finite groups have order that’s a power of 2. I don’t think this has been proved yet, but it’s obviously true empirically, and we also have a good idea as to why:

• Tom Leinster, Almost all of the first 50 billion groups have order 1024.

So, the map $\beta$ from the Burnside ring to the representation ring is surjective for most finite groups!

David Benson has listed the orders for which there are groups where $\beta$ is not surjective, and how many such groups there are of these orders:

24: 2,

40: 2,

48: 7,

56: 1,

60: 2,

72: 8,

80: 8,

84: 1,

88: 2,

96: 45,

104: 2,

112: 5,

120: 13,

132: 1,

136: 3,

140: 2,

144: 39,

152: 2,

156: 1,

160: 12,

168: 12,

171: 1,

176: 7,

180: 6,

184: 1,

192: 423,

200: 8,

204: 2,

208: 8,

216: 35,

220: 1,

224: 28,

228: 1,

232: 2,

240: 90,

248: 1,

252: 4,

260: 2,

264: 10,

272: 12,

276: 1,

280: 12,

288: 256,

296: 2,

300: 8,

304: 7,

308: 1,

312: 16,

320: 532,

328: 3,

333: 1,

336: 76,

340: 2,

342: 2,

344: 2,

348: 2,

352: 41,

360: 51,

364: 2,

368: 5,

372: 1,

376: 1,

380: 2.

He adds:

This seems to represent quite a small proportion of all finite groups, counted by order, even ignoring the $p$-groups.

All this is if we work over $\mathbb{Q}$. If we work over $\mathbb{C}$ the situation flip-flops, and I believe $\beta$ is usually not surjective. It already fails to be surjective for cyclic groups bigger than $\mathbb{Z}/2$!

Why? Because $\mathbb{Z}/n$ has a 1-dimensional representation where the generator acts as multiplication by a primitive $n$th root of unity, and since this is not a rational number when $n > 2$, this representation is not definable over $\mathbb{Q}$. Thus, one can show, there’s no way to get this representation as a formal difference of permutation representations, since those are always definable over $\mathbb{Q}$.

And this phenomenon of needing roots of unity to define representations is not special to cyclic groups: it happens for most finite groups as hinted at by Artin’s theorem on induced characters.

An example

Now, you’ll notice that I didn’t yet give an example of a finite group where the map $\beta$ from the Burnside ring to the representation ring fails to be surjective when we work over $\mathbb{Q}$.

In Serre’s book Linear Representations of Finite Groups, he asks us in Exercise 13.4 on page 105 to prove that $\mathbb{Z}/3 \times Q_8$ is such a group. Here $Q_8$ is the quaternion group, an 8-element subgroup of the invertible quaternions:

$$Q_8 = \{ \pm 1, \pm i , \pm j, \pm k \}$$

I couldn’t resist trying to understand why this is the counterexample Serre gave. First of all, $\mathbb{Z}/3 \times Q_8$ looks like a pretty weird group — why does it work, and why did Serre choose this one? Secondly, it has 24 elements, and I love the number 24. Thirdly, I love the quaternions.

Benson’s calculations show that the two smallest groups where $\beta$ is non-surjective have 24 elements. Drew Heard checked that these groups are $\mathbb{Z}/3 \times Q_8$ and the nontrivial semidirect product $\mathbb{Z}/3 \rtimes \mathbb{Z}/8$. But it’s still very interesting to do Serre’s exercise 13.4 and see why $\mathbb{Z}/3 \times Q_8$ is such a group.

To do this exercise, Serre asks us to use exercise 13.3. Here is my reformulation and solution of that problem. I believe any field $k$ characteristic zero would work for this theorem:

Theorem. Suppose $G$ is a finite group with a linear representation $\rho$ such that:

1. $\rho$ is irreducible and faithful

2. every subgroup of $G$ is normal

3. $\rho$ appears with multiplicity $n \ge 2$ in the regular representation of $G$.

Then the map from the Burnside ring of $G$ to the representation ring $R(G)$ of $G$ is not surjective.

Proof. It suffices to prove that the multiplicity of $\rho$ in any permutation representation of $G$ is a multiple of $n$, so that the class $[\rho] \in R(G)$ cannot be in the image of $R(G)$.

Since every finite $G$-set is a coproduct of transitive actions of $G$, which are isomorphic to actions on $G/H$ for subgroups $H$ of $G$, every permutation representation of $G$ is a direct sum of those on spaces of the form $k[G/H]$. (This is my notation for the vector space with basis $G/H$.) Thus, it suffices to show that the multiplicity of $\rho$ in the representation on $k[G/H]$ is $n$ if $H$ is the trivial group, and $0$ otherwise.

The former holds by assumption 3. For the latter, suppose $H$ is a nontrivial subgroup of $G$. Because $H$ is normal by assumption 2, every element $h \in H$ acts trivially on $k[G/H]$: we can see this by letting $h$ act on an arbitrary basis element $g H = H g \in G/H$:

$$h H g = H g .$$

Since $H$ is nontrivial, it contains elements $h \ne 1$ that act trivially on $k[G/H]$. But no $h \ne 1$ can act trivially on $\rho$ because $\rho$ is faithful, by assumption 1. Thus $\rho$ cannot be a subrepresentation of $k[G/H]$. That is, $\rho$ appears with multiplicity $0$ in $k[G/H]$.   ▮

Serre’s exercise 13.4 is to show the group $G = \mathbb{Z}/3 \times Q_8$ obeys the conditions of this theorem. As a hint, Serre suggests to embed $\mathbb{Z}/3$ and $Q_8$ in the multiplicative group of the algebra $\mathbb{H}_{\mathbb{Q}}$ (the quaternions defined over $\mathbb{Q}$). By letting $\mathbb{Z}/3$ act by left multiplication and $Q_8$ act by right multiplication, one obtains a 4-dimensional irreducible representation $\rho$ of $G$ which appears with multiplicity $n = 2$ in the regular representation. Furthermore $\rho$ is faithful and irreducible. Finally, every subgroup of $G$ is normal, because that’s true of $\mathbb{Z}/3$ and $Q_8$ — and since the orders of these groups are relatively prime, every subgroup of $G = \mathbb{Z}/3 \times Q_8$ is a product of a subgroup of $\mathbb{Z}/3$ and a subgroup of $Q_8$.

More work on the question

Alex Bartel wrote:

To add to what others have said: you may be interested in this paper “Rational representations and permutation representations of finite groups” of mine with Tim Dokchitser, which, among other things, contains an overview of what is (or was in 2016) known about the question, describes the algorithm that is used in the magma programme (implemented by Tim Dokchitser) that Drew Heard mentions in his answer, gives a closed form formula for the cokernel in quasi-elementary groups, and exhibits a family of finite simple groups with unbounded cokernel, in fact even unbounded exponent of cokernel (to our knowledge, in all finite simple groups for which one had known this cokernel up to that point, the cokernel was trivial).

The particular interest in quasi-elementary groups stems from the fact that one can reduce the study of this cokernel in arbitrary finite groups to that of its quasi-elementary subgroups and the induction-restriction maps between them. Observations of this nature have been made repeatedly over the years. In our paper we make this reduction very explicit.

Edit. In response to a question in a comment thread, and as an example of an interesting quasi-elementary group: the other group of order $24$ with non-trivial cokernel is $C_3\rtimes C_8$ with the unique non-trivial action. The cokernel of $\beta$ here comes about in the following way: there is an index $2$ subgroup $C_3\times C_4\cong C_{12}$. Take a faithful irreducible character of that, and induce up to the whole group. The resulting $2$-dimensional representation of $C_3\rtimes C_8$ has field of character values $\mathbb{Q}(i)$: the cube roots of unity have “disappeared” in the course of the induction. It turns out that the sum of this degree $2$ character and its Galois conjugate is realisable over $\mathbb{Q}$, but that it is not a virtual permutation representation. This is an illustration of Theorem 1.1 in my paper with Tim: $\rho$ is $4$-dimensional, $\hat\psi$ is the $2$-dimensional irreducible rational representation of $C_3$, the sum of the two faithful complex irreducible characters, and $\hat{\pi}$ is the unique $4$-dimensional irreducible rational representation of $C_8$, the sum of all its faithful complex irreducible characters.