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!

Group Homomorphism & Group Order

  1. Dec 15, 2014 #1
    I came across this problem in class note but I was stuck:

    Assume that ##G## be a group of order 21, assume also that ##G'## is a group of order 35, and let ##\phi## be a homomorphism from ##G## to ##G.'## Assume that ##G## does not have a normal subgroup of order 3. Show that ##\phi (g) = 1## for each element ##g## in ##G##.

    I know that most likely I need to use Lagrange's ##|G| = |G/H| \cdot |H|## and therefore ##|H|## divides ##|G|##, and also ##G/ker (\phi) \cong \phi (G)## and therefore ##|G/ker (\phi)| = |\phi (G)|##, but after those I am lost. A gentle and friendly explanation would therefore be very much appreciated.

    Thank you very much and happy holidays to all.
     
  2. jcsd
  3. Dec 15, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've already got everything you need. ##ker(\phi)=H## is a normal subgroup of ##G##. Therefore it has order 1,3,7 or 21 since those are the divisors of 21. ##\phi(G)## is a subgroup of ##G'## so it has order 1,5,7 or 35 since those are the divisors of 35. You are given that ##|ker(\phi)|## is not 3. Just try the other possibilities. I.e. if ##|ker(\phi)|=1## then ##|\phi(G)|=21/1=21##. Is that possible?
     
  4. Dec 15, 2014 #3
    Thank you very much! Now I get it.
     
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: Group Homomorphism & Group Order
  1. Group Homomorphism (Replies: 1)

  2. Homomorphism of groups (Replies: 2)

  3. Group Homomorphism? (Replies: 3)

Loading...