# Isomorphic simple groups

It is known that $SL_{4}(\mathbb{F}_2)\cong A_8$. Obviously, this is equivalent to the existence of a subgroup of $Sl_4(\mathbb{F}_2)$ of index $8$. How to find such a subgroup?

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Here's a construction of a 4-dimensional space over ${\bf F}_2$ together with an action of $A_7$ on it: math.harvard.edu/~elkies/Misc/A8.pdf – Noam D. Elkies Jan 9 '13 at 4:05
@Yu Just to satisfy my curiosity: why is it obvious that those two statements are equivalent? – MTS Jan 9 '13 at 6:10
@MTS: Suppose $H<SL_4(\mathbb F_2)$ has index $8$. Consider the left-multiplication action of $SL_4(\mathbb F_2)$ on the set of cosets $SL_4(\mathbb F_2)/H$ (which is a set of cardinality eight). The action is certainly transitive, and thus in particular is nontrivial. On the other hand, $SL_4(\mathbb F_2)$ is simple, so we get an injection of $SL_4(\mathbb F_2)$ into $S_8$. Counting cardinalities, we see it is of index two, but the only index two subgroup of $S_8$ is $A_8$ (since a subgroup of index two is normal and the abelianization of $S_8$ is $\mathbb Z/2$). – John Pardon Jan 9 '13 at 6:39
Note that ${\rm{SL}}_4/\mu_2 = {\rm{SO}}_6$ via the action of ${\rm{SL}}_4$ on the exterior square of its standard representation (with quadratic form $Q(\omega,\omega')=\omega\wedge\omega'$ valued in 4-forms). The fppf cohomology group ${\rm{H}}^1(\mathbf{F}_2,\mu_2)$ vanishes (Kummer theory), so the induced map ${\rm{SL}}_4(\mathbf{F}_2) \rightarrow {\rm{SO}}_6(\mathbf{F}_2)$ is bijective. Now use $S_8 \simeq {\rm{O}}_6(\mathbf{F}_2)$ defined by the 6-dimensional quadratic space $(V/(q|_V)^{\perp},q)$ where $V:=(\sum t_i=0)$ in affine 8-space over $\mathbf{F}_2$ and $q:= \sum_{i<j} t_i t_j$. – user29720 Jan 9 '13 at 7:00
Some of these comments would make fine answers. – S. Carnahan Jan 9 '13 at 7:46

Let $V\cong\mathbb F^3$ be a $3$-dim $\mathbb F_2$ space. Consider $V$ as a subgroup of $S(V)$. It is well-known that $N_{S(V)}(V) \cong V\rtimes GL(4)$. Then the isomorphism $A_8\cong GL_4$ because they are both even subgroups of $S_8=S(V)$.
Now the subgroup of index $8$ is the group $A_7$ which fixes the origin of $V$.
Edit: Apparently, I made a silly mistake along the way that $N_{S(V)}(V) \cong V\rtimes GL(4)$ (Should be $GL(3)$). Yet somehow I think it could be fixed and that the argument is somewhat equivalent to that of Elkies.