Conjugacy Action on Subsets is Group Action

Theorem

Let $\powerset G$ be the set of all subgroups of $G$.

For any $S \in \powerset G$ and for any $g \in G$, the conjugacy action:

$g * S := g \circ S \circ g^{-1}$

is a group action.

Proof

By Group Axiom $\text G 2$: Existence of Identity Element, $e \in G$, thus:

\(\ds e * S\)

\(=\)

\(\ds e \circ S \circ e^{-1}\)

Definition of $*$

\(\ds \)

\(=\)

\(\ds S\)

Definition of Identity Element

Thus Group Action Axiom $\text {GA} 2$ is seen to be fulfilled.

Then:

\(\ds \paren {g_1 \circ g_2} * S\)

\(=\)

\(\ds \paren {g_1 \circ g_2} \circ S \circ \paren {g_1 \circ g_2}^{-1}\)

Definition of $*$

\(\ds \)

\(=\)

\(\ds g_1 \circ g_2 \circ S \circ \paren {g_2^{-1} \circ g_1^{-1} }\)

Inverse of Group Product

\(\ds \)

\(=\)

\(\ds g_1 \circ \paren {g_2 \circ S \circ g_2^{-1} } \circ g_1^{-1}\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds g_1 * \paren {g_2 \circ S \circ g_2^{-1} }\)

Definition of $*$

\(\ds \)

\(=\)

\(\ds g_1 * \paren {g_2 * S}\)

Definition of $*$

Thus Group Action Axiom $\text {GA} 1$ is shown to be fulfilled.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.5$. Groups acting on sets: Example $105$