Automorphisms of subgroup of hamming cube under distance constraint - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:29:05Z http://mathoverflow.net/feeds/question/109635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109635/automorphisms-of-subgroup-of-hamming-cube-under-distance-constraint Automorphisms of subgroup of hamming cube under distance constraint rishig 2012-10-14T18:27:27Z 2013-06-07T18:22:00Z <p>Let \$S\$ be a subset of <code>\$\{0,1\}^n\$</code> such that any two elements of \$S\$ are at least (Hamming) distance 5 apart. I'm looking for an upper bound on the size of the automorphism group of \$S\$.</p> <p>There's a trivial upper bound of \$2^nn!\$ (the number of automorphisms of <code>\$\{0,1\}^n\$</code>), and an easy lower bound of \$2^{n/5}(n/5)!\$ (take S to be all elements of the form \$xxxxx\$, where \$x\$ is a bitstring of length \$n/5\$). </p> <p>Any bound of the form \$n!/n^{cn}\$ for any \$c>0\$ would be helpful.</p> http://mathoverflow.net/questions/109635/automorphisms-of-subgroup-of-hamming-cube-under-distance-constraint/109898#109898 Answer by David Bernier for Automorphisms of subgroup of hamming cube under distance constraint David Bernier 2012-10-17T11:49:25Z 2012-10-17T11:49:25Z <p>For the additive group {0, 1}^n , it seems to me that every non-singular binary nxn matrix provides one F_2 linear bijective map from {0, 1}^n to itself . As I recall, asymptotically the number of these bijective linear maps is at least C 2^(n^2), for some C>0 . In other words, a strictly positive proportion of random nxn matrices over F_2 has non-zero determinant, and an nxn matrix over F_2 has nxn entries, each being 0 or 1. When you say automorphism, are you referring to automorphisms of the additive group (F_2)^n ? Thanks. David Bernier</p>