Group Homomorphism & Group Order

  • #1
6
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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.
 

Answers and Replies

  • #2
Dick
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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.
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?
 
  • #3
6
0
Thank you very much! Now I get it.
 

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