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>