Decomposition of the symmetric group $\mathfrak{S}_{p+q}$

Question

Let $p,q \ge 1$ be two integers and $\mathfrak{S}_{p+q}$ the symmetric group of $\{1, \dots, p, p+1, \dots, p+q\}$. Denote:

• by $\Gamma_{p,q}$ the subgroup of permutations $\alpha \in \mathfrak{S}_{p+q}$ such that $\alpha(i)=i$ for all $i \in \{1, \dots, p\}$
• by $\Delta_{p,q}$ the subgroup of permutations $\beta \in \mathfrak{S}_{p+q}$ such that $\beta(j)=j$ for all $j \in \{p+1, \dots, p+q\}$
• and by $\mathcal{S}_{p,q}$ the set of permutations $\sigma \in \mathfrak{S}_{p+q}$ such that $\sigma(1) \lt \sigma(2) \lt \dots \lt \sigma(p)$ and $\sigma(p+1) \lt \sigma(p+2) \lt \dots \lt \sigma(p+q)$

I'm trying to prove that any $s \in \mathfrak{S}_{p+q}$ can be written in a unique way as $$s = \sigma \circ \alpha \circ \beta$$ where $\sigma \in \mathcal{S}_{p,q}$, $\alpha \in \Gamma_{p,q}$ and $\beta \in \Delta_{p,q}$.

I first tried to use a mockup with $p=2$, $q = 3$ and the permutation: $$s=\begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 4 & 2 & 5 & 1 & 3\end{pmatrix}$$ but I'm not able to get substantial things.

Thanks for your ideas!

Note: the topic is coming from the definition of the wedge product in exterior algebra.

Answer 1

I finally found a proof for the result of the question.

For any $s \in \mathfrak{S}_{p+q}$, let:

• $i_1 \lt \dots \lt i_q$ be such that $\{i_1, \dots, i_q\} = \{s(p+1), \dots, s(p+q)\}$, i.e. $i_k$ is the $k$-th element when $\{s(p+1), \dots, s(p+q)\}$ is ordered,
• and $j_1 \lt \dots \lt j_p$ with $\{j_1, \dots, j_p\} = \{s(1), \dots, s(p)\}$.

Then define:

• $\alpha \in \Gamma_{p,q}$ by $\alpha(i)=i$ for $i \in \{1, \dots, p\}$ and $\alpha^{-1}(p+k) = s^{-1}(i_k)$ for $k \in \{1, \dots, q\}$,
• $\beta \in \Delta_{p,q}$ by $\beta(j)=j$ for $j \in \{p+1, \dots, p+q\}$ and $\beta^{-1}(k) = s^{-1}(j_k)$ for $k \in \{1, \dots, p\}$,
• $\sigma = s \circ (\alpha \circ \beta)^{-1}$.

One can verify that:

• Any elements $a \in \Gamma_{p,q}$, $b \in \Delta_{p,q}$ commute.
• $s = \sigma \circ \alpha \circ \beta$,
• $\sigma$ belongs to $\mathcal{S}_{p,q}$ as for $k \in \{1, \dots, p\}$, $k = \beta \circ s^{-1}(j_k)$ and $\sigma(k) = j_k$, therefore $\sigma(1) \lt \dots \lt \sigma(p)$. We can have a similar argument to prove that $\sigma(p+1) \lt \dots \lt \sigma(p+q)$.

Let's now prove the uniqueness of the writing.

Suppose that

$$s = \sigma_1 \circ \alpha_1 \circ \beta_1 = \sigma_2 \circ \alpha_2 \circ \beta_2.$$

We have $\sigma_2 = \sigma_1 \circ \alpha \circ \beta$ with $\sigma_1, \sigma_2 \in \mathcal{S}_{p,q}$, $\alpha \in \Gamma_{p,q}$, $\beta \in \Delta_{p,q}$ and need to prove that $\alpha = \beta = \mathrm{id}$.

If $\beta \neq \mathrm{id}$, it exists $i\lt j \in \{1, \dots,p\}$ with $\beta(j) \lt \beta(i)$.

But then $$\sigma_2(j) = \sigma_1 \circ \beta(j) = \sigma_1(\beta(j)) \lt \sigma_1(\beta(i)) = \sigma_1 \circ \beta(i) = \sigma_2(i)$$ in contradiction with $\sigma_2 \in \mathcal{S}_{p,q}$. We can prove that $\alpha = \mathrm{id}$ with a similar argument. Finally $\alpha = \beta = \mathrm{id}$, which means $\alpha_1 = \alpha_2$ and $\beta_1 = \beta_2$ and therefore $\sigma_1 = \sigma_2$.