# Thread: phi: A --> G is a group homomorphism. Prove that a acts by the rule a.x=phi(a).x

1. ## phi: A --> G is a group homomorphism. Prove that a acts by the rule a.x=phi(a).x

Suppose that $G$ is a group acting on $X$ and that  $\phi : A \rightarrow G$ is a group homomorphism. Prove that $a$ acts on $X$ by the rule $a.x = \phi (a).x$.

2. Originally Posted by joestevens
Suppose that $G$ is a group acting on $X$ and that $\phi : A \rightarrow G$ is a group homomorphism. Prove that $a$ acts on $X$ by the rule $a.x = \phi (a).x$.
It pretty much falls out if you just follow the definition. What are you having trouble with?

3. Definition: $G$ acts on $X$ if there exists a function $\alpha : G \times X \rightarrow X$ such that
i) for $g,h \in G,\ \alpha_g \circ \alpha_h = \alpha_{gh}$
ii) $\alpha_e = e_X$.

Should I use i & ii above as the definition of the group action $G$ on $X$ and show $A$ also acts on $X$ by $a.x = \phi(a).x$?

Am I to take $A$ as a subgroup of $G$ or is $A$ just any group?

4. Originally Posted by joestevens
Definition: $G$ acts on $X$ if there exists a function $\alpha : G \times X \rightarrow X$ such that
i) for $g,h \in G,\ \alpha_g \circ \alpha_h = \alpha_{gh}$
ii) $\alpha_e = e_X$.

Should I use i & ii above as the definition of the group action $G$ on $X$ and show $A$ also acts on $X$ by $a.x = \phi(a).x$?

Am I to take $A$ as a subgroup of $G$ or is $A$ just any group?
Just use that definition. For example, i) is satisfied since $(ab)\cdot x=\phi(ab)x=\phi(a)\phi(b)x=a\cdot(\phi(b)x)=a\cdo t(b\cdot x)$