Group Homomorphism & Group Order

[rputra]

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.

 

[Dick]

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?

 

[rputra]

Thank you very much! Now I get it.