1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trivial group homomorphism from G to Q

  1. Jan 21, 2016 #1
    • Member warned about posting without the template
    Hello, I have to solve the following problem:

    Show that a homomorphism from a finite group G to Q, the additive group of rational numbers is trivial, so for every g of G, f(g) = 0.

    My work so far:

    f(x+y) = f(x)+f(y)
    I know that |G| = |ker(f)||Im(f)|

    I think that somehow I have to find that Im(f) = 1 but I don't know how. Can anybody help me please?

    Thanks in advance!
     
  2. jcsd
  3. Jan 21, 2016 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    You could use the fact that every element of G has a finite period.
     
  4. Jan 21, 2016 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The only finite subgroup of the additive group of rational numbers is {0}. (NOT {1}!)
     
  5. Jan 21, 2016 #4
    Okay I understand that, but now I only know that there is an isomorphism to Z/nZ with n the number of elements. Then I know that f is one-to-one, so |ker(f)| = 1 and Im(f) would equal n, but that's not what I need. Can you give me one more hint?
     
  6. Jan 21, 2016 #5

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Aren't you making this too complicated?
    Take an element ##g \in G##. Then ##g^p=e## for some positive integer ##p##, ( ##e## is the identity of G) . What information can you glean from ##f(g^p)=f(e)##?
     
  7. Jan 21, 2016 #6
    f(g^p) = f(e) so that is e of Q (= 0). And you can conclude that order(f(x)) is finite as well and then you can conclude that because {0} is the only finite subgroup of Q all elements of G will be mapped to 0?

    Thanks for the help!
     
  8. Jan 21, 2016 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I wasn't thinking that complicated! I would just use the fact that a homomorphism from group G to group H maps G to a subgroup of H.
     
  9. Jan 21, 2016 #8

    fresh_42

    Staff: Mentor

    Isn't this the same amount of work, sorry: line of proof? Because with that you have to prove your statement in #3.
     
  10. Jan 21, 2016 #9
    What are the requirements for f to be a group homomorphism?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trivial group homomorphism from G to Q
  1. Homomorphism Q (Replies: 5)

  2. Group Homomorphism? (Replies: 3)

Loading...