Group Action determines Bijection

Theorem

Let $*$ be a group action of $G$ on $X$.

Then each $g \in G$ determines a bijection $\phi_g: X \to X$ given by:

$\map {\phi_g} x = g * x$

Its inverse is:

$\phi_{g^{-1} }: X \to X$.

These bijection are sometimes called transformations of $X$.

Proof

Proof of Injectivity

Let $x, y \in X$

Then:

\(\ds \map {\phi_g} x\)

\(=\)

\(\ds \map {\phi_g} y\)

\(\ds \leadsto \ \ \)

\(\ds g * x\)

\(=\)

\(\ds g * y\)

\(\ds \leadsto \ \ \)

\(\ds g^{-1} * \paren {g * x}\)

\(=\)

\(\ds g^{-1} * \paren {g * y}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {g^{-1} g} * x\)

\(=\)

\(\ds \paren {g^{-1} g} * y\)

\(\ds \leadsto \ \ \)

\(\ds e * x\)

\(=\)

\(\ds e * y\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds y\)

Thus $\phi_g$ is an injection.

$\Box$

Proof of Surjectivity

Let $x \in X$.

Then:

\(\ds x\)

\(=\)

\(\ds e * x\)

\(\ds \)

\(=\)

\(\ds \paren {g g^{-1} } * x\)

\(\ds \)

\(=\)

\(\ds g * \paren {g^{-1} * x}\)

\(\ds \)

\(=\)

\(\ds \map {\phi_g} y\)

where $y = g^{-1} * x \in X$

Thus a group action is a surjection.

$\Box$

So a group action is an injection and a surjection and therefore a bijection.

$\blacksquare$

Also see

• Definition:Permutation Representation