- #1

Bashyboy

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- 5

## Homework Statement

Assume that ##G## is an abelian transitive subgroup of ##S_A## that acts on the set ##A##. Show that ##\sigma(a) \neq a## for all ##\sigma \in G - \{1\}## and all ##a \in A##. Deduce that ##|G| = |A|##.

## Homework Equations

A group is said to act transitively on a set if there is only one orbit.

Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{e\}$$

## The Attempt at a Solution

If ##G## acts transitively on ##A##, then by the above fact we see that the kernel of the action is faithful. This means that ##\sigma (a) = a## if and only if ##\sigma = e##, which is to say that ##\sigma(a) \neq a## for all ##\sigma \in G - \{e\}## and ##a \in A##. I am now going to switch from the function notation ##\sigma(a)## to the customary group action notation ##g \cdot a##.

Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g \cdot a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A = \cal{O}_a##. Clearly our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g \cdot a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g) \cdot a = a##. We will show this implies that ##(h^{-1}g)\cdot x = x## for all ##x \in A##. If ##x \in A##, then there exists a ##k \in G## such that ##x = k \cdot a##. Hence,

$$(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x$$

In the above we used commutativity several times. Since the kernel is trivial, ##h^{-1}g = e## or ##g = h##, thereby proving injectivity, and hence bijectivity.

Does this sound right?