Let $G$ be a finite group. The Burnside Ring $B(G)$ is defined as the Grothendieck group of the category $G$-set, The addition is induced by the disjoint union of $G$-sets. The multiplication on $B(G)$ is induced by the direct product of $G$-sets.

I am trying to prove that $B(G \times H) \cong B(G) \otimes B(H)$, where $G$ and $H$ are groups whose orders are relative primes.

I know that there is a map $B(G) \otimes B(H) \rightarrow B(G\times H)$ induced by the map $([X],[Y]) \mapsto [X\times Y]$, for a $G$-set $X$ and a $H$-set $Y$.

Now for a $(G\times H)$-set $Z$, the only map that I came up was the induced by $[Z] \mapsto ([Res_G^{G\times H}Z], [Res_H^{G \times H}Z])$.

I am stuck trying to prove that they are inverse of the each other and I don't know where the assumption about the orders play part.


Let $\mathsf{Set}(G)$ denote the set of isomorphism classes of finite $G$-sets. You extended the map

$$\mathsf{Set}(G)\times\mathsf{Set}(H)\ \longrightarrow\mathsf{Set}(G\times H) \tag{1}$$

to a ring homomorphism

$$ ~~~ B(G)\otimes B(H)\longrightarrow B(G\times H). \tag{2}$$

You seem to be proposing that there is an inverse to $(1)$,

$$ \mathsf{Set}(G)\times\mathsf{Set}(H)\longleftarrow\mathsf{Set}(G\times H) \tag{a} $$

which can be extended to an inverse to $(2)$,

$$ ~~~ B(G)\otimes B(H)\longleftarrow B(G\times H). \tag{b}$$

Your proposed map $(\textrm a)$ cannot be an inverse of $(1)$ because the map $(1)$ preserves cardinality, but the map $(\textrm a)$ does not. Indeed, $(1)$ does not have an inverse because it is not one-to-one. For example, given any trivial $G\times H$-set $Z$, if $|X\times Y|=|Z|$ and we interpret $X$ and $Y$ as trivial $G$-set and $H$-set respectively then the tuple $([X],[Y])$ gets mapped to $[Z]$. All of these tuples correspond to $|X|\cdot|Y|$ times the identity in the ring $B(G)\otimes B(H)$.

That the ring homomorphism is an isomorphism can be translated into purely $G\times H$-set-theoretic language, though. It is logically equivalent (exercise) to the statement

Proposition. Every $G\times H$-set $Z$ is uniquely expressible as sum of "factorizable" orbits,

$$ Z\cong \bigsqcup_{i=1}^r X_i\times Y_i, $$

i.e. where $X_1,\cdots,X_r$ are $G$-orbits and $Y_1,\cdots,Y_r$ are $H$-orbits.

(Nonuniqueness occurs when we omit the condition that the $X_i$s and $Y_i$s carry transitive actions.)

Proof of uniqueness. Every $G\times H$-set $Z$ is uniquely expressible as a disjoint union of orbits $Z_1\sqcup\cdots\sqcup Z_r$. "Factorizable" sets $X_i\times Y_i$ with $X_i$ a $G$-orbit and $Y_i$ an $H$-orbit are themselves $G\times H$-orbits. Conversely, assuming every $G\times H$-orbit is factorizable, we need to see orbits factor uniquely, i.e. that we can recover the $G$-set $X_i$ and $H$-set $Y_i$ from the $G\times H$-set $X_i\times Y_i$, up to isomorphism. But we can do that, since they are just the orbit spaces $X_i\cong (X_i\times Y_i)/H$ as a $G$-set and $Y_i\cong (X_i\times Y_i)/G$ as an $H$-set.

Proof of existence. We want any $G\times H$-orbit $Z_i$ to factor as $X_i\times Y_i$. Any $G\times H$-orbit is a coset space $(G\times H)/S$ (up to isomorphism) where $S\le G\times H$ is a subgroup. This is where the coprimality of $G$ and $H$ comes into play:

Lemma. If $|G|$ and $|H|$ are relatively prime, then any subgroup $S\le G\times H$ is itself a direct product of subgroups, $S=J\times K$ with $J\le G$, $K\le H$.

(proof. Project $S$ onto $G$ and $H$ and call the images $J$ and $K$. We know $|J|$ and $|K|$ are divisors of $|S|$, but since $|J|,|K|$ are coprime we can also say $|J|\cdot|K|$ is a divisor of $|S|$. But evidently $S\le J\times K$ in $G\times H$, so $S=J\times K$.)

Thus any $G\times H$-orbit, without loss of generality a coset space $(G\times H)/S$, can be factored into the form $(G\times H)/(J\times K)\cong (G/J)\times(H/K)$ as $G\times H$-sets, where $S=J\times K$.

  • $\begingroup$ Thanks! Wonderful argument. $\endgroup$ – C. Zhihao Nov 20 '17 at 18:12

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