Conjugacy Action
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Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Action on Group Elements
$G$ acts on itself by the rule $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$.
Also:
- $\operatorname{Stab} \left({x}\right) = C_G \left({x}\right)$, where $C_G \left({x}\right)$ is the centralizer of $x$ in $G$.
- $\operatorname{Orb} \left({x}\right) = C_{x}$, where $C_{x}$ is the conjugacy class of $x$.
Action on Subgroups
Let $X$ be the set of all subgroups of $G$.
For any $H \le G$ and for any $g \in G$, we define: $\forall g \in G, H \in X: g * H = g \circ H \circ g^{-1}$
This is a group action.
Also:
- $\operatorname{Stab} \left({H}\right) = N_G \left({H}\right)$ where $N_G \left({H}\right)$ is the normalizer of $H$ in $G$.
- $\operatorname{Orb} \left({H}\right)$ is the set of subgroups conjugate to $H$.
Proof
Elements
- Clearly GA-1 is fulfilled as $e * x = x$.
- GA-2 is shown to be fulfilled thus:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({g_1 \circ g_2}\right) * x\) | \(=\) | \(\displaystyle \left({g_1 \circ g_2}\right) \circ x \circ \left({g_1 \circ g_2}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 \circ g_2 \circ x \circ g_2^{-1} \circ g_1^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 * \left({g_2 \circ x \circ g_2^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 * \left({g_2 * x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
- $\operatorname{Stab} \left({x}\right) = C_G \left({x}\right)$ follows from the definition of centralizer: $C_G \left({x}\right) = \left\{{g \in G: g \circ x = x \circ g}\right\}$.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle z\) | \(\in\) | \(\displaystyle \operatorname{Stab} \left({x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle z\) | \(\in\) | \(\displaystyle \left\{ {g \in G: g \circ x \circ g^{-1} = x}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle z\) | \(\in\) | \(\displaystyle \left\{ {g \in G: g \circ x = x \circ g}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Furthermore, since the powers of $x$ commute with $x$, $\left \langle {x} \right \rangle \in C_G \left({x}\right)$.
- $\operatorname{Orb} \left({x}\right) = C_{x}$ follows from the definition of the conjugacy class.
$\blacksquare$
Subgroups
- Clearly GA-1 is fulfilled as $e * H = H$.
- GA-2 is shown to be fulfilled thus:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({g_1 \circ g_2}\right) * H\) | \(=\) | \(\displaystyle \left({g_1 \circ g_2}\right) \circ H \circ \left({g_1 \circ g_2}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 \circ g_2 \circ H \circ g_2^{-1} \circ g_1^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 * \left({g_2 \circ H \circ g_2^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle g_1 * \left({g_2 * H}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
- $\operatorname{Stab} \left({H}\right) = \left\{{g \in G: g \circ H \circ g^{-1} = H}\right\}$ which is how the normalizer is defined.
- $\operatorname{Orb} \left({H}\right) = \left\{{g \circ H \circ g^{-1}: g \in G}\right\}$ from the definition.
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.5$: Example $103$
- John F. Humphreys: A Course in Group Theory (1996): $\S 10$: Examples $10.5, \ 10.7, \ 10.10$