Decomposition of permutations and wedge products. Let $V$ be an $\mathbb{R}$-vector space. Denote the space of all alternating $k$-linear forms from $V^k$ to $\mathbb{R}$ by ${\cal A}_k(V, \mathbb{R})$
Suppose $f\in{\cal A}_p(V, \mathbb{R})$ and $g\in{\cal A}_q(V, \mathbb{R})$. Munkres (Analysis on Manifolds) defines the wedge product of $f$ and $g$, $f\wedge g \in {\cal A}_{p+q}(V, \mathbb{R})$, as an alternating $(p+q)$-form given by:
$$
(f\wedge g)(\mathbf{x}) = \cfrac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)})
$$
where $x_i$ is the $i^{th}$ component of $\mathbf{x}$ and $\epsilon(\sigma)$ is the sign of the permutation.
In my differential geometry course, the instructor defined wedge product as:
$$(f\wedge g)(\mathbf{x}) = \sum_{\sigma\in S_{p,q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)})$$
where $S_{p,q} = \{ \sigma \in S_{p+q} : \sigma(1) < \sigma(2) < \dots < \sigma(p)$ and $\sigma(p+1) < \sigma(p+2) < \dots < \sigma(p+q) \}$.
How do I show the equivalence of these two definitions? Here is my attempt:
First of all, for convenience, we define the following subsets of $S_{p+q}$.
$ P = \{\sigma \in S_{p+q} : \sigma\ \ \text{fixes}\ \  p+1, p+2, \dots, p+q\}$
edit: $P$ is just a copy of $S_p$ in $S_{p+q}$
$ Q = \{\sigma \in S_{p+q} : \sigma\ \ \text{fixes}\ \  1, 2, \dots, p\}$
edit: $Q$ is just a copy of $S_q$ in $S_{p+q}$
(By "$\sigma$ fixes $i$", I mean that $\sigma(i) = i$).
We know that, $|S_{p, q}| = {{p+q}\choose{p}}$.
Further, I want to claim that given any $\sigma \in S_{p+q}$, we can decompose, $\sigma = \phi \rho \tau$, where $\phi \in S_{p, q}$, $\rho \in P$ and $\tau \in Q$ (This is something that I believe to be true, but couldn't quite prove it).
Assuming this fact, we show the equivalence as follows: (The intermediate steps make use of the fact that $\rho$  and $\tau$ are disjoint and hence commute and also that $f$ and $g$ are alternating maps).
$$
\begin{align}
&(f\wedge g)(\mathbf{x})\\
& = \cfrac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau)f(x_{\phi\rho\tau(1)}, x_{\phi\rho\tau(2)}, \dots, x_{\phi\rho\tau(p)})g(x_{\phi\rho\tau(p+1)}, x_{\phi\rho\tau(p+2)}, \dots, x_{\phi\rho\tau(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau)f(x_{\phi\tau(1)}, x_{\phi\tau(2)}, \dots, x_{\phi\tau(p)})g(x_{\phi\rho(p+1)}, x_{\phi\rho(p+2)}, \dots, x_{\phi\rho(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau) \epsilon(\tau) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)})\epsilon(\rho)g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi) \epsilon(\tau)^2 \epsilon(\rho)^2 f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\
& = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}} p!q! \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\
& = \sum_{\phi \in S_{p, q}} \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})
\end{align}
$$
which completes the proof.
Now, the only thing that remains to be shown is that the decomposition $\sigma = \phi\rho\tau$ is actually possible. But I am not sure how to do that. Any hints will be appreciated.
 A: Notations
I use $k, l$ instead of $p, q$.
I use $v_i$ instead of $x_i$.
Lastly, I use $\omega, \eta$ instead of $f, g$.

Define the equivalence relation $\sim$ on $S_{k+l}$ by setting $\sigma \sim \sigma'$ iff
$$\{\sigma(1), \ldots, \sigma(k)\} = \{\sigma'(1), \ldots, \sigma'(k)\}.$$
(Note that the equality is of sets.)
It is easily checked that $\sim$ is indeed an equivalence relation. Moreover, if $\sigma \sim \sigma',$ then we also have
$$\{\sigma(k+1), \ldots, \sigma(k+l)\} = \{\sigma'(k+1), \ldots, \sigma'(k+l)\}.$$
Let $[\sigma]$ denote the equivalence class of $\sigma.$
We make the following simple observations:

*

*Each equivalence class contains the same number of elements.

*The above number is $k!l!.$

*Every equivalence class contains exactly one $(k, l)$ shuffle.

*Every $(k, l)$ shuffle is contained in some equivalence class.

Now, if we show that the quantity $\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})$ is the same for all $\sigma$ belonging to a fixed equivalence class, then we would be done.
That is because, we could simply choose the shuffle present in the equivalence class as the representative of the class and then the two expressions would coincide. To see this better, let $\Pi_1, \ldots \Pi_r$ denote the distinct equivalence classes and let $\sigma_i \in \Pi_i$ be the shuffle in that class. Then, we have
$$S_{k+l} = \bigsqcup_{i=1}^r \Pi_i$$
and thus,
\begin{align} 
  & \dfrac{1}{k!l!}\sum_{\sigma \in S_{k+l}}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\
  =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}\sum_{\sigma \in \Pi_i}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\
  =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}\sum_{\sigma \in \Pi_i}\operatorname{sgn}(\sigma_i)\omega(v_{\sigma_i(1)}, \ldots, v_{\sigma_i(k)})\eta(v_{\sigma_i(k+1)}, \ldots, v_{\sigma_i(k+l)})\\
  &\text{note that now the inner quantity is independent of $\sigma$}\\
  =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}(k!l!)\operatorname{sgn}(\sigma_i)\omega(v_{\sigma_i(1)}, \ldots, v_{\sigma_i(k)})\eta(v_{\sigma_i(k+1)}, \ldots, v_{\sigma_i(k+l)})\\
  =&\;\sum_{\sigma \in S_{(k, l)}}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}).
\end{align}
Thus, now all we need to finish is the following claim.

Claim. If $[\sigma] = [\sigma'],$ then
\begin{align} 
\operatorname{sgn}(\sigma)&\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\
=& \operatorname{sgn}(\sigma')\omega(v_{\sigma'(1)}, \ldots, v_{\sigma'(k)})\eta(v_{\sigma'(k+1)}, \ldots, v_{\sigma'(k+l)}).
\end{align}
Proof. Since $\{\sigma(1), \ldots, \sigma(k)\} = \{\sigma'(1),\ldots, \sigma'(k)\},$ we can find a permutation $\tau \in S_{k+l}$ such that
$$\tau\sigma(i) = \sigma'(i), \quad i = 1, \ldots, k$$
and $\tau$ acts as identity on $\{\sigma(k+1), \ldots, \sigma(k+l)\}.$
Similarly, we can find a permutation $\pi \in S_{k+l}$ such that
$$\pi\sigma(i) = \sigma'(i), \quad i = k+1, \ldots, k+l$$
and $\pi$ acts as identity on $\{\sigma(1), \ldots, \sigma(k)\}.$
Thus, we actually get
$$\pi\tau\sigma(i) = \sigma'(i), \quad i = 1, \ldots, k+l.$$
That is to say, $\sigma' = \pi\tau\sigma.$
In particular, we have $\operatorname{sgn}\sigma' = \operatorname{sgn}\pi\cdot\operatorname{sgn}\tau\cdot\operatorname{sgn}\sigma.$
This also gives us that
$$\operatorname{sgn}\sigma'\cdot\operatorname{sgn}\pi\cdot\operatorname{sgn}\tau = \operatorname{sgn}\sigma.$$
With that in place, we prove the claim via the following calculation.
\begin{align} 
  & \operatorname{sgn}\sigma'\cdot\omega(v_{\sigma'(1)}, \ldots, v_{\sigma'(k)})\eta(v_{\sigma'(k+1)}, \ldots, v_{\sigma'(k+l)})\\
  =&\;\operatorname{sgn}\sigma'\cdot\omega(v_{\tau\sigma(1)}, \ldots, v_{\tau\sigma(k)})\eta(v_{\pi\sigma(k+1)}, \ldots, v_{\pi\sigma(k+l)})\\
  & \text{Now, we use that $\omega$ and $\eta$ are alternating}\\
  =&\;\operatorname{sgn}\sigma'\cdot\operatorname{sgn}\tau\cdot\operatorname{sgn}\pi\cdot\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\
  =&\;\operatorname{sgn}\sigma\cdot\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}) & \blacksquare
\end{align}
A: That decomposition is indeed possible, that is, given a permutation $\sigma \in S_{p+q}$, we can write $\sigma = \phi \rho \tau$,  where,

*

*$\phi$ is a $(p, q)$-shuffle

*$\rho$ is a permutation of first $p$ symbols (i.e., it acts as identity on the last $q$ symbols)

*$\tau$ is a permutation on the last $q$ symbols (i.e., it acts as identity on the first $p$ symbols).

I will copy some steps from @Aryaman's answer. We define an equivalence on $S_{p+q}$ as follows:
$$\sigma \sim \sigma' \quad \text{iff.} \quad \{\sigma(1), \dots, \sigma(p)\} = 
\{\sigma'(1), \dots, \sigma'(p)\}$$
Now, each equivalence class has a unique $(p, q)$-shuffle, because, a $(p, q)$-shuffle is uniquely determined by the image of $\{1, \dots, p\}$ under the shuffle. That is, two distinct $(p, q)$-shuffles cannot give the same image of the first $p$ symbols.
Now, given any permutation $\sigma \in S_{p+q}$, suppose the unique $(p,q)$-shuffle in the class of $\sigma$ is $\phi$. This means that $$ \{\sigma(1), \dots, \sigma(p)\} = \{\phi(1), \dots, \phi(p)\}.$$
Next, we find a permutation $\rho$ such that $\sigma(i) = \phi\rho(i)$ for $i = 1,\dots, p$, and $\rho(i) = i$ for $i = p+1, \dots, p+q$.
(Note that, this $\rho$ permutes only the first $p$ symbols and fixes the remaining ones.)
Similarly, we find a permutation $\tau$ such that $\sigma(i) = \phi\tau(i)$ for $i = p+1,\dots, p+q$, and $\tau(i) = i$ for $i = 1, \dots, p$.
This gives us the required decomposition, $\sigma = \phi\rho\tau$.
