Results 1 to 4 of 4

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

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    42

    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 $\displaystyle a.x = \phi (a).x$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    Quote Originally Posted by joestevens View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    Posts
    42
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    Quote Originally Posted by joestevens View Post
    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 $\displaystyle (ab)\cdot x=\phi(ab)x=\phi(a)\phi(b)x=a\cdot(\phi(b)x)=a\cdo t(b\cdot x)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Nov 17th 2011, 12:27 PM
  2. Group homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Nov 11th 2011, 04:45 PM
  3. group homomorphism
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 22nd 2010, 04:51 AM
  4. Prove group homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 29th 2010, 09:02 AM
  5. group homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Feb 22nd 2009, 09:54 AM

Search Tags


/mathhelpforum @mathhelpforum