Is every finite group a proper quotient of a finite primitive group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:11:52Z http://mathoverflow.net/feeds/question/57723 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57723/is-every-finite-group-a-proper-quotient-of-a-finite-primitive-group Is every finite group a proper quotient of a finite primitive group? Colin Reid 2011-03-07T19:59:38Z 2011-03-07T21:54:31Z <p>Let \$G\$ be a finite group. Is there necessarily a finite primitive permutation group \$P\$ and a normal subgroup \$N>1\$ of \$P\$ such that \$P/N \cong G\$?</p> <p>If not, what restrictions are there on quotients of finite primitive permutation groups?</p> http://mathoverflow.net/questions/57723/is-every-finite-group-a-proper-quotient-of-a-finite-primitive-group/57737#57737 Answer by Derek Holt for Is every finite group a proper quotient of a finite primitive group? Derek Holt 2011-03-07T21:54:31Z 2011-03-07T21:54:31Z <p>Yes. We can assume that \$G\$ is a transitive permutation group. Let \$S\$ be any primitive finite simple group, such as \$A_5\$ in its natural representation. Now let \$P\$ be the wreath product of \$S\$ with \$G\$ using the product action, which has degree \$d(P) = d(S)^{d(G)}\$. This gives a primitive group, and the quotient of \$P\$ with the base group \$S^{d(G)}\$ of the wreath product is isomorphic to \$G\$. </p> <p>Note that the primitive wreath product action of \$S \wr G\$ can also be described as its action by multiplication on the cosets of its maximal subgroup \$T \wr G\$, where \$T\$ is a point stabilizer in \$S\$.</p>