Homomorphism and subgroups

Question

Let $G$, $H$ be groups and let $L$ be a subgroup to $H$. Show that if $h: G → H$ is a homomorphism then $K = \{x\in G\,|\,h(x)\in L\}$ is a subgroup to $G$.

Hi everyone, I'm not sure how to draw the connection between the homomorphism and the subgroup $K$. How do I prove this?

Thanks for all the help

Answer 1

Since $f$ is a homomorphism, $f(e_G)=e_H\in L$. Therefore, $e_G\in K$.

If $k_1,k_2\in K$, then $f(k_1k_2)=f(k_1)f(k_2)\in L$. Therefore, $k_1,k_2\in K$.

Finally, if $k\in K$, then $f(k)f(k^{-1})=f(kk^{-1})=f(e_G)=e_H$ and therefore $f(k^{-1})=f(k)^{-1}\in L$. So, $k^{-1}\in K$.

Answer 2

The image of a subgroup is certainly non-empty. Moreover, if $y_1,y_2\in K$, then there are $x_1,x_2\in L$ such that $f(x_i)=y_i$ for all $i$. Now, $y_1y_2^{-1}=f(x_1)(f(x_2))^{-1}$.

$f(x_1)(f(x_2))^{-1}$ is the only $\alpha\in H$ such that $\alpha f(x_2)=f(x_1)$. But $f(x_1x_2^{-1})$ satisfies this property too, because $f(x_1x_2^{-1})f(x_2)=f(x_1x_2^{-1}x_2)=f(x_1)$ and $x_1x_2^{-1}\in L$ because $L$ is a subgroup. So $y_1y_2^{-1}=f(x_1x_2^{-1})\in K$ for all $y_1,y_2\in K$. Therefore, $K$ is a subgroup.