Conjugacy Action is not Transitive

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Theorem

Let $\struct {G, \circ}$ be a non-trivial group whose identity is $e$.

Let $*: G \times G \to G$ be the conjugacy group action:

$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$


Then $*$ is not a transitive group action.


Proof

Proof by Counterexample:

For $G$ to be a transitive group action, the orbit of any element of $G$ needs to be the whole of $G$.

Take $h = e$.

Then:

\(\ds \forall g \in G: \, \) \(\ds g * e\) \(=\) \(\ds g \circ e \circ g^{-1}\)
\(\ds \) \(=\) \(\ds g \circ g^{-1}\) Group Axiom $\text G 2$: Existence of Identity Element
\(\ds \) \(=\) \(\ds e\) Group Axiom $\text G 3$: Existence of Inverse Element

Thus by definition of orbit:

$\Orb e = \set e$

Only when $G$ is the trivial group, that is: $G = \set e$, does $\Orb e = G$.

Hence the result by definition of transitive group action.

$\blacksquare$


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