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

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

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

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

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

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

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

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