Tensor products of exterior powers For my advanced linear algebra course, I have to prove that $(\wedge^{k}V \otimes V \otimes V) \cap (V \otimes \wedge^{k}V \otimes V) = \wedge^{k+1} V \otimes V$, regarded as subspaces of $V^{\otimes k+2}$, and for $k \geq 2$. Intuitively this seems like it should be true, since being antisymmetric in the first $k$ places, and being antisymmetric in the 'second to $k+1$'st places implies being anti-symmetric in the first $k+1$ places, right? I do not know how to prove this formally though. 
 A: Let's assume that your $V$ is a vector space over a field (I guess the result still holds when $V$ is a flat module over a commutative ring, but I don't have the time to try and prove it), so $V$ has a basis $\left(e_1,e_2,...,e_n\right)$ (note that I'm assuming it finite purely for the ease of notation). Thus, the vector space $V^{\otimes k+1}$ has a basis $\left(e_{i_1}\otimes e_{i_2}\otimes ...\otimes e_{i_{k+1}}\right)_{\left(i_1,i_2,...,i_{k+1}\right)\in\left\lbrace 1,2,...,n\right\rbrace^{k+1}}$. We will be working with respect to this basis (so "coordinates" will always mean coordinates with respect to this basis).
We need to show that $\left(A^k V \otimes V\otimes V\right)\cap \left(V\otimes A^k V\otimes V\right) = A^{k+1} V\otimes V$, where $A^i V$ denotes the subspace of $V^{\otimes i}$ consisting of all antisymmetric tensors. It is known that whenever $P$ and $Q$ are vector spaces, $P_1$ and $P_2$ are vector subspaces of $P$, and $Q_1$ and $Q_2$ are vector subspaces of $Q$, we have $\left(P_1\otimes Q_1\right)\cap\left(P_2\otimes Q_2\right)=\left(P_1\cap P_2\right)\otimes\left(Q_1\cap Q_2\right)$. Applied to $P = V^{\otimes k+1}$, $Q = V$, $P_1 = A^k V \otimes V$, $P_2 = V \otimes A^k V$, $Q_1 = V$, $Q_2 = V$, this yields
$\left(A^k V \otimes V \otimes V\right) \cap \left(V \otimes A^k V \otimes V\right) = \left(\left(A^k V\otimes V\right) \cap \left(V\otimes A^k\right)\right) \otimes \underbrace{\left(V\cap V\right)}_{=V}$
$= \left(\left(A^k V\otimes V\right) \cap \left(V\otimes A^k\right)\right) \otimes V$.
Hence, it is enough to show that
$\left(A^k V\otimes V\right) \cap \left(V\otimes A^k\right) = A^{k+1} V$.
The $\supseteq $ inclusion of this equality is obvious, so we only need to prove the $\subseteq$ inclusion. So let $f\in \left(A^k V \otimes V\otimes V\right)\cap \left(V\otimes A^k V\otimes V\right)$. We need to show that $f \in A^{k+1} V\otimes V$. In order to prove this, we must check that
1. for any $j\in\left\lbrace 1,2,...,k\right\rbrace $ and any $\left(k+1\right)$-tuple $\left(i_1,i_2,...,i_{k+1}\right)\in\left\lbrace 1,2,...,n\right\rbrace^{k+1}$ such that $i_j = i_{j+1}$, the 
$\left(e_{i_1}\otimes e_{i_2}\otimes ...\otimes e_{i_{k+1}}\right)$-coordinate of $f$ is $0$.
2. for any $j\in\left\lbrace 1,2,...,k\right\rbrace $ and any $\left(k+1\right)$-tuple $\left(i_1,i_2,...,i_{k+1}\right)\in\left\lbrace 1,2,...,n\right\rbrace^{k+1}$, the
$\left(e_{i_1}\otimes e_{i_2}\otimes ...\otimes e_{i_{k+1}}\right)$-coordinate of $f$ equals minus the $\left(e_{i_1}\otimes e_{i_2}\otimes ...\otimes e_{i_{j-1}} \otimes e_{i_{j+1}} \otimes e_{i_j} \otimes e_{i_{j+2}} \otimes e_{i_{j+3}} \otimes ... \otimes e_{i_{k+1}}\right)$-coordinate of $f$.
(Depending on your definition of an antisymmetric tensor, you might need to think about why it is enough to prove 1. and 2.. Hint: the symmetric group is generated by transpositions of adjacent integers.)
But both of the properties 1. and 2. are easy to show: when $j\leq k-1$, one has to use $f\in A^k V\otimes V$, whereas for $j = k$ one needs to apply $f \in V\otimes A^k V$.
A: Let $W$ be a representation of $G$ over a field $K$ and define a homomorphism $\chi:G\to K^\times$. Define the "twisted subspace" $W^{H,\chi}:=\{w\in W:hw=\chi(h)w~\forall h\in H\}$ (if $\chi$ is the trivial homomorphism, this is precisely the subspace of $H$-invariants). Then

Lemma. $W^{H,\chi}\cap W^{K,\chi}=W^{\langle H\cup K\rangle,\chi}$. Proof. The inclusion $\supseteq$ should be clear automatically, since we have $A\ge B\implies W^{A,\chi}\le W^{B,\chi}$ and $\langle H\cup K\rangle\ge H,K$. For the reverse inclusion, we need to show that for any word $\sigma=h_1k_1\cdots h_rk_r$, that $\sigma v=\chi(\sigma)v$, given $v\in W^{H,\chi}\cap W^{K,\chi}$; this should be clear by working on the word one letter at a time.

The reason I include this is that it allows us to formulate the problem in a way amenable to doing some deductions and I feel the higher generality makes the problem look easier.
In your case, $S_{k+2}$ acts on $W=V^{\otimes(k+2)}$, and you want $$W^{S_k\times1\times1,\epsilon}\cap W^{1\times S_k\times1,\epsilon}=W^{S_{k+1}\times1,\epsilon},$$ where $\epsilon$ is the sign homomorphism. Thus you want to show that the subgroups $S_k\times1$ and $1\times S_k$ generate all of the permutations $S_{k+1}$ ultimately (we are ignoring the final tensor factor).
