"Natural" choice of representative of left cosets in a symmetric group

Question

Let $n$ be a positive integer and let $n=n_1+\cdots+ n_k$ be a partition of $n$, where each $n_i$ is a positive integer. Let $\Sigma_n$ be the symmetric group on $n$ letters. Then $$\Sigma_{n_1}\times\cdots\times\Sigma_{n_k}$$ is a subgroup of $\Sigma_n$. My question is:

Is there a "natural" choice of representatives of left cosets of $\Sigma_{n_1}\times\cdots\times\Sigma_{n_k}$ in $\Sigma_n$?

Answer 1

tl;dr: No

I tried to begin by making the word "natural" precise. Let $\mathsf{C}$ be the following category:

• An object is a pair $(X,P)$ where $X$ is a finite set and $P \subseteq 2^X$ is a partition of $X$.
• A morphism $(X,P) \to (Y,P')$ is an injective function $f : X \to Y$ such that, for all $S \in P$, there exists $T \in P'$ such that $f(S) \subseteq T$.

Let $\Gamma : \mathsf{C} \to \mathsf{Set}$ be the functor defined as follows:

• Given an object $(X, \{X_1, \dots, X_n\})$, $\Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ can be canonically identified with a subgroup of $\Sigma_X$ via the inclusions $X_i \subset X$, and this subgroup does not depend on the choice of ordering of the $X_i$'s. Thus, it makes sense to define $$\Gamma(X, \{X_1, \dots, X_n\}) = \Sigma_X/(\Sigma_{X_1} \times \dots \times \Sigma_{X_n} )$$ (this is just the set of left cosets, not necessarily a group).
• Given a morphism $f : (X,\{X_1, \dots, X_n\}) \to (Y,\{Y_1, \dots, Y_m\})$, the injective function $f : X \to Y$ defines a map $f_* : \Sigma_X \to \Sigma_Y$ by $$f_*(\sigma)(y) = \begin{cases} y &: y \notin f(X) \\ f(\sigma(f^{-1}(y))) &: y \in f(X) \end{cases}$$ This map has the property that $f_*(\Sigma_{X_1} \times \dots \times \Sigma_{X_n}) \subseteq \Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}$. This means that if $\sigma_1^{-1} \sigma_2 \in \Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ then $f_*(\sigma_1)^{-1} f_*(\sigma_2) \in \Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}$, so it makes sense to define $$\Gamma(f)(\sigma (\Sigma_{X_1} \times \dots \times \Sigma_{X_n})) = f_*(\sigma) (\Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}).$$

For any $(X,P) \in \mathsf{C}$, the identity function on $X$ gives a morphism $\varepsilon_{(X,P)} : (X,\{\{x\} : x \in X\}) \to (X,P)$. This defines a natural transformation $\varepsilon : D \to \operatorname{id}_{\mathsf{C}}$, where $D : \mathsf{C} \to \mathsf{C}$ is the functor $(X,P) \mapsto (X,\{\{x\} : x \in X\})$ (as the notation may suggest, $\varepsilon$ is the counit of an adjunction).

Now consider the natural transformation $\Gamma \varepsilon : \Gamma \circ D \to \Gamma$. For any $(X,P) \in \mathsf{C}$, we have that $$\Gamma(D(X,P)) = \Sigma(X,\{\{x\} : x \in X\}) = \Sigma_X \left/ \prod_{x \in X} \Sigma_{\{x\}}\right. = \Sigma_X / \{e\}$$ is naturally isomorphic to $\Sigma_X$. By canonically identifying $\Sigma_X / \{e\}$ and $\Sigma_X$, we get that $(\Gamma \varepsilon)_{(X,P)}$ is precisely the quotient map $$\Sigma_X \to \Gamma(X,P).$$ A splitting of this quotient map is precisely a choice of representatives for the cosets, so to look for a natural choice of representatives, we could ask: does there exist a natural splitting of this map? In other words:

Question Does there exist a natural transformation $\xi : \Gamma \to \Gamma \circ D$ such that $(\Gamma \varepsilon) \circ \xi = \operatorname{id}_\Gamma$?

Answer No.

Proof. Suppose such a natural transformation $\xi$ exists. Let $X$ be a set with more than one element. Let $A = (X,\{\{x\} : x \in X\})$ and $B = (X,\{X\})$. By naturality, we get the following commutative diagram:

$\require{AMScd}$ \begin{CD} \Gamma(A) @>{\Gamma(\varepsilon_B)}>> \Gamma(B)\\ @V{\xi_A}VV @V{\xi_B}VV\\ \Sigma_X @>{\Gamma(D(\varepsilon_B)) = \operatorname{id}_{\Sigma_X}}>> \Sigma_X \end{CD}

Since $\Gamma(B)$ is a singleton set, $\xi_A = \operatorname{id}_{\Sigma_X} \circ \xi_A = \xi_B \circ \Gamma(\varepsilon_B)$ is constant. Then $\operatorname{id}_{\Gamma(A)} = \Gamma(\varepsilon_A) \circ \xi_A$ is constant, but $\Gamma(A) \cong \Sigma_X$! This contradicts the fact that $X$ has more than one element.

Maybe this is too restrictive – why do we demand that $\xi$ be natural with respect to all morphisms in $\mathsf{C}$? A seemingly reasonable but weaker condition would be to ask that $\xi$ is natural with respect to isomorphisms in $\mathsf{C}$ (formally, we can form the subcategory $\mathsf{C}' \leq \mathsf{C}$ which contains only isomorphisms, then restrict everything down to $\mathsf{C}'$). However, even this condition (which is about as little as one could hope for) cannot be satisfied.

Proof. Suppose there is a splitting $\xi$ of $\Gamma \varepsilon$ which is natural with respect to isomorphisms in $\mathsf{C}$. Let $Q = (\{1, \dots, 6\}, \{\{1,2,3\},\{4,5,6\}\})$, let $G = \Sigma_{\{1, \dots, 6\}}$ and let $H = \Sigma_{\{1,2,3\}} \times \Sigma_{\{4,5,6\}} \leq G$. Let $N$ be the normalizer of $H$. By a simple computation, $N = \operatorname{Aut}_{\mathsf{C}}(Q)$, and the induced action of $N$ on $\Gamma(Q) = G/H$ is given by conjugation. The naturality of $\xi$ says that that $\xi_Q : G/H \to G$ is $N$-equivariant (where $N$ is also acting by conjugation on $G$). We now restrict our attention to $\xi_Q|_{N/H} : N/H \to N$. First of all, this is still a morphism of $N$-sets, because $N/H$ and $N$ are closed under conjugation by elements of $N$. By direct computation, $[N : H] = 2$, so $N/H$ is an abelian group. But then the conjugation action of $N$ on $N/H$ is just given by $$n \cdot (n' H) = (nn'n^{-1})H = (nH)(n'H)(nH)^{-1} = n'H,$$ so this action is trivial! The $N$-equivariance of $\xi_Q|_{N/H}$ then tells us that $\xi_Q(N/H) \subseteq Z(N)$. By direct computation, $N$ has trivial center, which means $\xi_Q|_{N/H}$ is constant. But $\xi_Q$ is injective, and $\lvert N/H \rvert > 1$, so this is impossible.

I make no claim that this is the simplest counterexample! Anyway, I think this argument shows that there can't be anything like a natural choice of representatives for the cosets of $\Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ in $\Sigma_X$.