Conjugacy Action is not Transitive
Jump to navigation
Jump to search
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
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.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(v)}$