Isomorphic simple groups

2013-01-09T03:58:53Z

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?

Answer by Quang Hoang for Isomorphic simple groups

2013-01-10T18:14:53Z

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 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.

Isomorphic simple groups - MathOverflow

Isomorphic simple groups

Answer by Quang Hoang for Isomorphic simple groups