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