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\begin{center}
\begin{large}
The isomorphism GL$_4(\Ft)\cong A_8$ (outline)
\end{large}
\\
Noam D. Elkies
\end{center}

The simple groups $A_8$ and GL$_4(\Ft)$ both have
$\frac12 8! = 15\cdot14\cdot12\cdot8 = 20160$ elements.
One might then expect that they are isomorphic.  This
in fact turns out to be the case.  One way to show this
is to construct a \hbox{4-dimensional} vector space~$V$ over $\Ft$ on
which $A_7$ acts nontrivially by linear transformations;
this will embed $A_7$ as an index-8 subgroup
of GL$_4(\Ft)$, and thus a homomorphism of GL$_4(\Ft)$ into $S_8$.
Since GL$_4(\Ft)$ is simple, this homomorphism will have trivial
kernel, and image contained in $A_8$.  It will thus yield an
embedding of GL$_4(\Ft)$ into $A_8$, which must then be an isomorphism
because the group orders coincide.

Let $H\subset S_7$ be the automorphism group of the projective plane
$\Pi_2$.  We know that $\#H=168$.  Moreover, $H$ is simple, so
$H\subset A_7$.  Thus $H$ has $[A_7:H]=\frac12 7!/168=15$ cosets
in~$A_7$.  To these 15 we will add {\bf0} to obtain~$V\!$, and show
that $V$ has the structure of an $\Ft$-vector space.  That is, we
will define for each $P,Q\in V$ a natural element $P+Q\in V$\/
and show that this addition law satisfies the axioms of an $\Ft$-vector
space.

The cosets of~$H$ in $S_7$ are the 30 different ways to identify an
unstructured 7-element set~$\Sigma$ with the points of~$\Pi_2$.  We'll
show that two of these fall in the same $A_7$-orbit if and only if
they have exactly one line in common.  (Otherwise they share either
three lines or none, but we will not need this.)  We do this by
noting that given one $\Pi_2$ structure~$P$ on~$\Sigma$ and one of
its seven lines~$l$, there are two other $\Pi_2$ structures in
the same $A_7$-orbit whose only common line with~$P$ is~$l$, namely
the images of~$P$ under $g,g^2$ where $g\in A_7$
% is a 3-cycle disjoint from~$l$.
%   [that works, but the following is probably easier to see...]
is a 3-cycle permuting~$l$.
This is enough because it accounts for all $1+2\cdot7=15$
structures in the orbit of~$P$.  We now define $P+Q$ for $P,Q\neq0$
as follows: if $P=Q$ then of course $P+Q=0$; else $P,Q$ have a unique
line~$l$ in common, and $P+Q$ is the third $\Pi_2$-structure in the
same $A_7$ containing~$l$.  The only obstruction to verifying that
this makes $V$ into an $\Ft$-vector space is checking that the
identity $(P+Q)+R = P+(Q+R)$ for distinct $P,Q,R\in V-\{{\bf0}\}$
without a common line.

Fortunately $A_7$ acts (simply) transitively on the
$15\cdot14\cdot12=2520$ such triples $(P,Q,R)$, so it is
enough to check the identity on any one of them.  To prove
the transitivity it suffices to verify that if $g\in A_7$ fixes
each of $P,Q,R$ then $g$ is the identity.  But then $g$ fixes
each of the lines common to two of $P,Q,R$, which are
three {3-element} subsets of~$\Sigma$ any two of which
meet at a point.  They thus separate $\Sigma$ into singletons,
and so $g$ cannot move any element of~$\Sigma$.  [Note that
it follows in particular that the action of $A_7$ on $V-\{{\bf0}\}$
is doubly transitive.]  Confirming that $(P+Q)+R = P+(Q+R)$
for some appropriate $P,Q,R$\/ is straightforward.
This completes the identification of GL$_4(\Ft)$ with $A_8$.

Alternatively we could have obtained $V$ directly as a
representation of $A_8$, by identifying the nonzero points
with orbits of AGL$_3(\Ft)\subset A_8$, i.e.\ with half
of the 30 ways of arranging an \hbox{8-element} set into a 3-(8,4,1)
Steiner system.  That makes it somewhat trickier to define
the linear structure on~$V\!$, though.

We can even recover $S_8$ from GL$_4(\Ft)$.  Recall that
$S_8={\rm Aut}(A_8)$, so we need only find an outer automorphism
of GL$_4(\Ft)$.  But we already know such an automorphism:
the ``contragredient'' involution $A\leftrightarrow{}^{\!t}A^{-1}$.
\end{document}
