Prove or disprove if $f:G\to K$ is group homomorphism then necessarily $|kerf|\le |imf|$

Question

I wrote a counterexample: $$G=(\Bbb Z_5,+_{mod 5})=\{0,1,2,3,4\}$$ $$K=(\Bbb Z_3,+_{mod 3})=\{0,1,2\}$$ $$e_k=0,f(a)=3a$$ so: $kerf=\{0,1,2,3,4\}$ and $imf=\{0\}$

my counterexample is right? $$$$Thanks.

Answer 1

Yeah, your counterexample looks good.

In general if you have groups $G$ and $H$ you can consider the morphism $\varphi: G\rightarrow H$ defined by $\varphi(g)=e_H$ for all $g\in G$.

The image is $\{1_H\}$ and the kernel is $G$. So this is a valid counterexample whenever $G$ is not trivial.