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

There are groups of order $2^n \cdot p$ where $\beta$ is not surjective, but David Benson listed how many of them there are for various orders, and there are rather few:

“Most” finite groups are $p$-groups, for which the map is an isomorphism. But quite a lot of groups of order $2^n \cdot p$ are examples where it isn’t. There are 45 such groups of order 96, for example.

Here is a list of the first few orders for which there are such groups, and how many there are:

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.

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 from the Burnside ring to the representation ring fails to be surjective when we work over $\mathbb{Q}$. I recently had a lot of fun doing an exercise in Serre’s book Linear Representations of Finite Groups, where he asks us in Exercise 13.4 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.

Right now I believe $\mathbb{Z}/3 \times Q_8$ is the smallest group for which $\beta : A(G) \to R(G)$ is non-surjective. I’ve asked on MathOverflow, but so far nobody has answered that question. I got a lot of useful information about my other question, though: is $\beta$ surjective for most finite groups, or not?

To solve Serre’s problem 13.4, he asked you 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 $Q_8$.