An abelian group has the same cardinality as any sets on which it acts transitively

Suppose G \leq S_A is an abelian and transitive subgroup. Show that \sigma(a) \neq a for all \sigma \neq 1 in G and all a \in A. Deduce that |G| = |A|.


For each a \in A, we have 1 = \bigcap_{\sigma \in G} \sigma \mathsf{stab}(a) \sigma^{-1} = \bigcap_{\sigma \in G} \sigma \sigma^{-1} \mathsf{stab}(a) = \bigcap_{\sigma \in G} \mathsf{stab}(a) = \mathsf{stab}(a) since G is abelian. So for all a \in A and \sigma \neq 1, we have \sigma(a) \neq a.

Because the action of G is transitive, if we fix a \in A then for every b \in A there exists \sigma \in A such that b = \sigma \cdot a. Suppose now that \sigma \cdot a = \tau \cdot a; then \tau^{-1} \sigma \cdot a = a, so that \tau^{-1}\sigma \in \mathsf{stab}(a) = 1. Hence \sigma = \tau. Thus we can say, for a fixed a \in A, that for every b \in A there exists a unique \sigma \in G such that \sigma \cdot a = b. In this way we define an injective mapping \varphi_a : A \rightarrow G. Hence |A| \leq |G|.

Now with a \in A fixed, we define \psi_a : G \rightarrow A by \psi_a(\sigma) = \sigma \cdot a. This mapping is injective since if \sigma \cdot a = \tau \cdot a, we have \tau^{-1} \sigma \in \mathsf{stab}(a) = 1. Thus |G| \leq |A|.

Hence |G| = |A|.

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