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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An elementary answer in terms of symmetric groups. Let $V\cong\mathbb F^3$ be a $3$dim $\mathbb F_2$ space. Consider $V$ as a subgroup of $S(V)$. It is wellknown 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. 

