Permutation groups: bases and moved points
Let [tex]G[/tex] be a finite group which acts (on the right) faithfully and transitively on a set [tex]\Omega[/tex].
Call "base" of this action a subset [tex]\Gamma[/tex] of [tex]\Omega[/tex] such that if an element [tex]g \in G[/tex] fixes every element of [tex]\Gamma[/tex] it is the identity - in other words [tex]\{g \in G\ |\ \alpha g=\alpha\ \forall \alpha \in \Gamma\}=\{1\}[/tex] -, and whose cardinality is the minimum among the cardinalities of the subsets of [tex]\Omega[/tex] with this property. Let [tex]b(G)[/tex] denote this minimum cardinality.
Given [tex]g \in G[/tex] let [tex]\text{supp}(g)[/tex] be the set of the elements of [tex]\Omega[/tex] moved by [tex]g[/tex], in other words [tex]\text{supp}(g):=\{\alpha \in \Omega\ |\ \alpha g \neq \alpha\}[/tex]. Let [tex]\mu(G)[/tex] be the minimum of the [tex]|\text{supp}(g)|[/tex] when [tex]g[/tex] varies in [tex]G-\{1\}[/tex].
1. Compute [tex]b(G)[/tex] and [tex]\mu(G)[/tex] when [tex]G = C_n = \langle (1 ... n) \rangle[/tex] (the cyclic group of order [tex]n[/tex]), [tex]G=S_n[/tex] (the symmetric group on [tex]n[/tex] objects, that is the group of bijective maps [tex]\{1,...,n\} \to \{1,...,n\}[/tex]), [tex]G=A_n[/tex] (the alternating group on [tex]n[/tex] objects) acting in the usual way on [tex]\{1,...,n\}[/tex] and [tex]G=\text{GL}(m,q)[/tex] (the group of the [tex]\mathbb{F}_q[/tex]-linear automorphisms of [tex]{\mathbb{F}_q}^m[/tex]) acting in the usual way on [tex]{\mathbb{F}_q}^m-\{0\}[/tex].
2. Prove that [tex]\mu(G) \cdot b(G) \geq |\Omega|[/tex].
Point 1 is easy, it is useful just to get used to the definitions. You will appreciate the fact that the name "base" in this context is compatible with the concept of base in linear algebra: the bases of [tex]GL(V)[/tex] (the group of the linear automorphisms of [tex]V[/tex]) acting on [tex]V-\{0\}[/tex] are exactly the bases of [tex]V[/tex] in the sense of linear algebra, so that [tex]b(GL(V))=\dim(V)[/tex].
Point 2 is very tricky!
Call "base" of this action a subset [tex]\Gamma[/tex] of [tex]\Omega[/tex] such that if an element [tex]g \in G[/tex] fixes every element of [tex]\Gamma[/tex] it is the identity - in other words [tex]\{g \in G\ |\ \alpha g=\alpha\ \forall \alpha \in \Gamma\}=\{1\}[/tex] -, and whose cardinality is the minimum among the cardinalities of the subsets of [tex]\Omega[/tex] with this property. Let [tex]b(G)[/tex] denote this minimum cardinality.
Given [tex]g \in G[/tex] let [tex]\text{supp}(g)[/tex] be the set of the elements of [tex]\Omega[/tex] moved by [tex]g[/tex], in other words [tex]\text{supp}(g):=\{\alpha \in \Omega\ |\ \alpha g \neq \alpha\}[/tex]. Let [tex]\mu(G)[/tex] be the minimum of the [tex]|\text{supp}(g)|[/tex] when [tex]g[/tex] varies in [tex]G-\{1\}[/tex].
1. Compute [tex]b(G)[/tex] and [tex]\mu(G)[/tex] when [tex]G = C_n = \langle (1 ... n) \rangle[/tex] (the cyclic group of order [tex]n[/tex]), [tex]G=S_n[/tex] (the symmetric group on [tex]n[/tex] objects, that is the group of bijective maps [tex]\{1,...,n\} \to \{1,...,n\}[/tex]), [tex]G=A_n[/tex] (the alternating group on [tex]n[/tex] objects) acting in the usual way on [tex]\{1,...,n\}[/tex] and [tex]G=\text{GL}(m,q)[/tex] (the group of the [tex]\mathbb{F}_q[/tex]-linear automorphisms of [tex]{\mathbb{F}_q}^m[/tex]) acting in the usual way on [tex]{\mathbb{F}_q}^m-\{0\}[/tex].
2. Prove that [tex]\mu(G) \cdot b(G) \geq |\Omega|[/tex].
Point 1 is easy, it is useful just to get used to the definitions. You will appreciate the fact that the name "base" in this context is compatible with the concept of base in linear algebra: the bases of [tex]GL(V)[/tex] (the group of the linear automorphisms of [tex]V[/tex]) acting on [tex]V-\{0\}[/tex] are exactly the bases of [tex]V[/tex] in the sense of linear algebra, so that [tex]b(GL(V))=\dim(V)[/tex].
Point 2 is very tricky!
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