Conjugacy Action on Identity

Theorem

Let $G$ be a group whose identity is $e$.

For the conjugacy action:

$\order {\Orb e} = 1$

and thus:

$\Stab e = G$

Proof

\(\ds g * e\)

\(=\)

\(\ds g e g^{-1}\)

\(\ds \)

\(=\)

\(\ds g g^{-1}\)

\(\ds \)

\(=\)

\(\ds e\)

So the only conjugate of $e$ is $e$ itself.

Thus:

$\Orb e = \set e$

and so:

$\order {\Orb e} = 1$

From the Orbit-Stabilizer Theorem, it follows immediately that:

$\Stab e = G$

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.5$. Groups acting on sets: Example $103$