Exercise at the Beginning of Part II in Fulton's Book on Young Tableaux In Fulton's Book Young Tableaux, there's an Exercise at the beginning of part II for which I cannot find a solution (there doesn't seem to be one for this exercise in my copy of the book). It reads: 

Exercise. Show that, if $e_1,\ldots,e_m$ is a basis for the ($\mathbb{C}$-vectorspace) $E$, then the images of the vectors $(e_i\wedge e_j)\otimes e_k$, for all $i<j$ and $i\le k$, form a basis of \[ E^{(2,1)} := \left.{\textstyle\bigwedge^2E}\otimes E\middle/\left((u\wedge v)\otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\:\middle|\:u,v,w\in E\right)\right.. \]

First, I do not see why, for $i<j<k$, the equality \[(e_i\wedge e_j)\otimes e_k = (e_k\wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j\] already holds in $\bigwedge^2 E\otimes E$. But even that aside, I do not see how to solve the exercise. Any help would be welcome.
 A: It is enough to show two things:
1) The listed element generate the space.
We know that $(e_i \wedge e_j) \otimes e_k$ with no restraints on the indices generate the space, and we can clearly restrict to $i < j$ by the antisymmetric behaviour of the wedge product. Now, if $i > k$, we use the identity:
$$((u∧v)\otimes w=(w∧v)⊗u+(u∧w)⊗v$$
for $u=e_i$, $v=e_j$ and $w=e_k$.
This gives
$$((e_i∧e_j)\otimes e_k=(e_k∧e_j)⊗e_i+(e_i∧e_k)⊗e_j= (e_k∧e_j)⊗e_i-(e_k∧e_i)⊗e_j,$$
which shows that the vector is actually a linear combination of two of our given vectors. Therefore, they are indeed a generating set.
2) The listed elements are independent.
Note that the relations in the ideal are multilinear in $u$,$v$,$w$, so it is enough to use the relations for the basis vectors to generate the full ideal.
Now, suppose that you have a linear combination of the given vectors that gives zero in the quotient space. This means that the linear combination lies in the ideal which means that the linear combination is a linear combination of relations in the ideal.
You can now look at the smallest index of an $e_i$ that occurs in this expression and check that its coefficient is 0.
A: You should know that the $e_i \wedge e_j (i<j)$ are a basis of $\bigwedge^2E$, so that after tensoring you have a basis
$$
e_i \wedge e_j \otimes e_k \quad(1\leq i< j \leq n,1\leq k \leq n)
$$
for $\bigwedge^2E \otimes E$.
All you have to show is that the elements where $k<i$ are in the subspace $V$ generated by
$$
(u\wedge v) \otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\
$$
but look at 
$$
(e_i \wedge e_j ) \otimes e_k \equiv (e_k \wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j \quad (\text{mod }  V)
$$
and other permutations of the indices. Assume $k<i<j$. Can you show that the left side is $\equiv 0$ mod $V$?
EDIT: this is not an answer, the necessary computations are more difficult than I expected!!
