- #1

- 35

- 0

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.