For homomorphism $\phi: G\to H$, if $G_{1}\leq G$ and $|G_{1}|=n$, then $|\phi(G_{1})|$ divides $n$.

Question

Could someone please help me with this proof? I am not sure how to show this...

For homomorphism $\phi: G\to H$, if $G_{1}\leq G$ and $|G_{1}|=n$, then $|\phi(G_{1})|$ divides $n$.

My work so far: By definition of $H$, $\phi(G_{1}) \subset H$. Since $\phi$ is a homomorphism that sends a group to a group, $\phi(G_{1})$ is a group. Then $\phi(G_{1})\leq H$.

I want to use properties of $\phi$ but I am not sure what I can say...

Answer 1

You may think in the following way:

By Lagrange Theorem, $|\phi(G_1)|= |G_1/(G_1\cap\ker \phi)| $ divids $G_1$.

I hope this suffices to give you the idea.

$:\,)$