Isomorphic simple groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:41:49Z http://mathoverflow.net/feeds/question/118417 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118417/isomorphic-simple-groups Isomorphic simple groups Yu 2013-01-09T03:58:53Z 2013-01-10T18:29:54Z <p>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? </p> http://mathoverflow.net/questions/118417/isomorphic-simple-groups/118559#118559 Answer by Quang Hoang for Isomorphic simple groups Quang Hoang 2013-01-10T18:14:53Z 2013-01-10T18:29:54Z <p>An elementary answer in terms of symmetric groups. </p> <p>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)\$. </p> <p>Now the subgroup of index \$8\$ is the group \$A_7\$ which fixes the origin of \$V\$. </p> <p>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.</p>