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I am trying to prove that $\operatorname{GL}(4,2)\cong \operatorname{Alt}(8)$. As part of the proof I already know that $\operatorname{Alt}(7)\subset \operatorname{GL}(4,2)$ and that $[\operatorname{GL}(4,2):\operatorname{Alt}(7)]=8$. The proof then says that since $\operatorname{GL}(4,2)$ is simple, the isomorphism must hold, but I cannot see why! Any help would be greatly appreciated.

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up vote 7 down vote accepted

You've shown that $GL(4,2)$ has a subgroup $H$ of index $8$. $GL(4,2)$ acts on the left cosets of H by left multiplication. This defines a homomorphism $f$ from $GL(4,2)$ to $S_8$. Since $GL(4,2)$ is simple and the action is nontrivial, the kernel of $f$ is trivial, so $f$ is actually an embedding of $GL(4,2)$ into $S_8$. Thus $GL(4,2)$ is isomorphic to an index $2$ subgroup of $S_8$, of which $A_8$ is the only one.

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