----- > [!proposition] Proposition. ([[homomorphism induced by group action]]) > Let $G$ be a [[group]] [[group action|acting on]] a set $X$: $\begin{align} G \times X \to X \\ (g,x) \mapsto g \cdot x. \end{align}$ Then we have a [[group homomorphism|homomorphism]] $\begin{align} G \xrightarrow{\phi} \text{Perm}(X) \\ g \mapsto (\cdot)_{g}, \end{align}$ where $(\cdot)_{g} \in \text{Perm}(X)$ is defined by $(\cdot )_{g}(x):= g \cdot x$. Moreover, if the [[group action]] is [[faithful group action|faithful]], then $\phi$ is an [[injection]]. > [!proof]- Proof. ([[homomorphism induced by group action]]) > Let's verify that $\phi$ is a [[group homomorphism|homomorphism]]. Write $\begin{align} \phi(g_{1} g_{2})(x)= & (\cdot)_{g_{1}g_{2}}(x) \\ = & (g_{1}g_{2}) \cdot x \\ = & g_{1} \cdot (g_{2} \cdot x) \\ = & g_{1} \cdot (\cdot)_{g_{2}}(x) \\ = & (\cdot) _{g_{1}} \circ (\cdot)_{g_{2}}(x) \\ = & \phi(g_{1}) \circ \phi(g_{2})(x), \end{align}$ which is the result. In general, $(\cdot)_{g}=\id$ if and only if $gx=x$ for all $x \in G$. Therefore, if we further suppose the [[group action|action]] is [[faithful group action|faithful]], then the only element $g$ for which $\phi(g)=(\cdot )_{g}=\id$ is $g=e$. Hence $\ker \phi=(e)$ and so [[group homomorphism is injective iff kernel is trivial|injectivity of]] $\phi$ holds. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.ta[](group%20homomorphism%20is%20injective%20iff%20kernel%20is%20trivial%20iff%20is%20a%20monomorphism.md)--- tags: - proposition - algebra/group-theory Created: 2023-10-02 Modified: 2023-10-02 mathLink: auto --- > [!metadata]- > Created::[[2023-10-02]] Modified::[[2023-10-02]] [[Tags]]:: #proposition Noteworthy Uses:: *[[Noteworthy Uses]]* Proved By:: *[[Proved By|Crucial Dependencies]]* Intuition:: *[[Intuition]]* ----- > [!proposition] Proposition. ([[homomorphism induced by group action]]) > Let $G$ be a [[group]] [[group action|acting on]] a set $X$: $\begin{align} G \times X \to X \\ (g,x) \mapsto g \cdot x. \end{align}$ Then we have a [[group homomorphism|homomorphism]] $\begin{align} G \xrightarrow{\phi} \text{Perm}(X) \\ g \mapsto (\cdot)_{g}, \end{align}$ where $(\cdot)_{g} \in \text{Perm}(X)$ is defined by $(\cdot )_{g}(x):= g \cdot x$. Moreover, if the [[group action]] is [[faithful group action|faithful]], then $\phi$ is an [[injection]]. > [!proof]- Proof. ([[homomorphism induced by group action]]) > Let's verify that $\phi$ is a [[group homomorphism|homomorphism]]. Write $\begin{align} \phi(g_{1} g_{2})(x)= & (\cdot)_{g_{1}g_{2}}(x) \\ = & (g_{1}g_{2}) \cdot x \\ = & g_{1} \cdot (g_{2} \cdot x) \\ = & g_{1} \cdot (\cdot)_{g_{2}}(x) \\ = & (\cdot) _{g_{1}} \circ (\cdot)_{g_{2}}(x) \\ = & \phi(g_{1}) \circ \phi(g_{2})(x), \end{align}$ which is the result. In general, $(\cdot)_{g}=\id$ if and only if $gx=x$ for all $x \in G$. Therefore, if we further suppose the [[group action|action]] is [[faithful group action|faithful]], then the only element $g$ for which $\phi(g)=(\cdot )_{g}=\id$ is $g=e$. Hence $\ker \phi=(e)$ and so [[group homomorphism is injective iff kernel is trivial|injectivity of]] $\phi$ holds. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```