I stated myself the following claim:
Let $f:G\rightarrow G'$ a morphism of groups, then $card(Im(f))$ divides $card(G)$ and $card(G')$
I wanted to proof this. We just know that we have two functions $$G->>Im(f)\hookrightarrow G'$$ where $->>$ is a surjection and $\hookrightarrow$ is an injection. Since f is a morphism $Im(f)$ is a subgroup of $G'$ and therefore $card(Im(f))\,|\,card(G')$. Now I don't know how to show that it also divides $card(G)$. Is there a possibility to show this?
Thank you for your help
Take $I=\mathrm{Im}f$, $K=\ker f$. Then we know $G/K$ is isomorph to $I$. $G/K$ groups all elements of $G$ into disjoint categories of size $|K|$, so $$ |G| = |K||G/K| = |K||I|$$