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 $P/N \cong G$?
If not, what restrictions are there on quotients of finite primitive permutation groups?
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$\begingroup$ seems intersting (the question and the answer) but what does "primitive" mean ? $\endgroup$– JoëlCommented Mar 8, 2011 at 2:49
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$\begingroup$ @Joël: it means that the action does not preserve any non-trivial equivalence relations on the set being permuted. $\endgroup$– Colin ReidCommented Mar 8, 2011 at 15:05
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$\begingroup$ An equivalent condition is that a permutation group $G$ is primitive if and only if it is transitive and a point stabilizer is a maximal subgroup of $G$. $\endgroup$– Derek HoltCommented Mar 9, 2011 at 9:34
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1 Answer
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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 $G$.
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 $S$.
answered Mar 7, 2011 at 21:54
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$\begingroup$ Excellent, that's just what I was hoping for. $\endgroup$ Commented Mar 7, 2011 at 22:32