Conjugacy Action on Subgroups is Group Action

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Theorem

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


Let $X$ be the set of all subgroups of $G$.

For any $H \le G$ and for any $g \in G$, the conjugacy action:

$g * H := g \circ H \circ g^{-1}$

is a group action.


Proof

Clearly GA-1 is fulfilled as $e * H = H$.

GA-2 is shown to be fulfilled thus:

\(\displaystyle \paren {g_1 \circ g_2} * H\) \(=\) \(\displaystyle \paren {g_1 \circ g_2} \circ H \circ \paren {g_1 \circ g_2}^{-1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1 \circ g_2 \circ H \circ g_2^{-1} \circ g_1^{-1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1 * \paren {g_2 \circ H \circ g_2^{-1} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1 * \paren {g_2 * H}\) $\quad$ $\quad$

$\blacksquare$


Also see


Sources