TITLE: $GL_n(\mathbb{F})$ contains a copy of $\mathbb{F}^{n-1}$
QUESTION [7 upvotes]: It is a fact of matrix multiplication that
$$\left(
\begin{matrix}
1 & a & b \\&1&\\&&1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & a' & b'\\&1&\\&&1
\end{matrix}
\right)
=
\left(
\begin{matrix}
1 & a +a'&b+b' \\&1&\\&&1
\end{matrix}
\right).
$$
Therefore $GL_n(\mathbb{F})$ contains a copy of the abelian group $\mathbb{F}^{n-1}$. Can we show that it does or does not contain a copy of $\mathbb{F}^n$, or any higher power?
Edit: Derek points out that, for even $n$, $\frac {n^2}{4}$ and for odd $n$, $\frac{n^2-1}{4}$ are possible. Can we show this the is highest value which is possible for all fields (while of course for specific fields higher may be possible)?

REPLY [5 votes]: There is a copy of $F^{(\lfloor \frac{n}{2} \rfloor)^2}$ in ${\rm GL}_n(F)$ for all $n$: put the nonzero off-diagonal entries in a square in the top right corner of the matrix.
I believe that, at least for finite fields, and possibly for ${\mathbb Q}$, this is the largest possible for $n>3$.
For some fields, such as ${\mathbb R}$, the additive groups of $F^n$ are isomorphic for all $n>0$.