What is a more formal name for the wedgey group? The rule $(v_1,w_1)⋅(v_2,w_2)=(v_1+v_2,w_1+w_2+(v_1∧v_2))$ defines a group structure on the vector space $V⊕(V∧V)$ whenever $V$ is itself a vector space over some field $F$.
What is a more common name for this group?


*

*The group is nilpotent of class at most $2$, with commutator $[(v_1,w_1),(v_2,w_2)] = (0,2(v_1∧v_2))$.

*The group has exponent $\operatorname{char}(F)$ since $(v_1,w_1)^n = (v_1^n,w_1^n)$.

*When $\operatorname{char}(F) = 2$, this is just an elementary abelian $2$-group.

*When $\dim(V) = 1$, this is just the abelian group $V$, so has a faithful $2$-dimensional F-module.

*When $\dim(V) = 2$ and $\operatorname{char}(F)≠2$, this is a maximal unipotent subgroup of $\operatorname{GL}(3,F)$, so has a faithful $3$-dimensional $F$-module.


I don't recognize it when $\dim(V) = 3$.

When $\dim(V) = 3$ and $\operatorname{char}(F)≠2$, does the group have a faithful $F$-module of dimension independent of $F$?

Is this a maximal unipotent subgroup of some classical group or does it otherwise have a standard description as a matrix group?
I am primarily concerned with the case that $F$ is a finite prime field (and am hoping to do better than a permutation module of dimension $|F|^5$), but I assume many sorts of answers should be mostly independent of the field.
I don't have much hope for large $\dim(V)$, since the nilpotency class of the group remains at $2$, while maximal unipotent subgroups should increase their nilpotency class with their rank.  Maybe $\dim(V)=3$ is small enough though.
 A: I think you have a nil-2 product of $n$ copies of $F$ (rather than two copies of $V$, as I originally wrote in comments). 
If $A_1,\ldots,A_n$ are abelian groups, then their 2-nilpotent product is their free product modulo the second term of the lower central series of the free product (in general, you take the free product and mod out by the intersection of the second term of the lower central series with the cartesian, the kernel of the map from the free to the direct product).
For the abelian case, the elements of the 2-nil product can be written uniquely as
$$a_1\cdots a_n\gamma$$
where $a_i\in A_i$, and $\gamma$ is in the commutator subgroup. The commutator subgroup is isomorphic to
$$\sum_{1\leq i\lt j\leq n} A_j\otimes A_i.$$
(In the general case, you take the abelianizations). 
The multiplication rule is
$$(a_1\cdots a_n\gamma)(b_1\cdots b_n\delta) = (a_1b_1\cdots a_nb_n)\left(\gamma+\delta+\sum_{1\leq i\lt j\leq n}b_i\otimes a_j\right).$$
If we take each $A_i$ isomorphic to $F$, then the commutator subgroup will be isomorphic to an $F$-vector space of dimension $\binom{n}{2}$, with basis given by $v_{ji}$, where $1\leq i\lt j\leq n$ (identifying $1_j\otimes 1_i$ with $v_{ji}$). If you make the "convention" that $v_{ij} = -v_{ji}$ and $v_{ii}=0$, then you get exactly the wedge $F^n\wedge F^n$. 
I am a little worried about that factor of $2$ in the commutator; there would be no problem with defining the $2$-nil product of groups of exponent $2$ and getting a 2-nil group (of exponent 4), which you cannot do with your set up. Indeed, if I think of this as the $2$-nilpotent product of $n$-copies of $F$, with elements expressed as sums of multiples of $\mathbf{1}_1,\ldots,\mathbf{1}_n$, I would write:
\begin{align*}
[\alpha_1\mathbf{1}_1+\cdots+\alpha_n\mathbf{1}_n,\beta_1\mathbf{1}_1+\cdots+\beta_n\mathbf{1}_n] &= \prod_{i,j=1}^n [\alpha_i\mathbf{1}_i,\beta_j\mathbf{1}_j]\\
&= \prod_{1\leq i\lt j\leq n} [\mathbf{1}_j,\mathbf{1}_i]^{\alpha_j\beta_i-\alpha_i\beta_j}.
\end{align*}
On the other hand, you would have
\begin{align*}
2(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n)\wedge(\beta_1\mathbf{v}_1 + \cdots + \beta_n\mathbf{v}_n) &=
2\sum_{i,j=1}^n \alpha_i\mathbf{v}_i\wedge\beta_j\mathbf{v}_j\\
&= 2\sum_{i,j=1}^n (\alpha_i\beta_j)\mathbf{v}_i\wedge\mathbf{v}_j\\
&= 2\sum_{1\leq i\lt j\leq n} (\alpha_j\beta_i - \beta_j\alpha_i)\mathbf{v}_i\wedge\mathbf{v}_j,
\end{align*}
that is, twice as much as I do. 
