I have the following problem:
Let $G$ be a group and $H<G$ be a subgroup. Let $E=\{(g,k)\in G\times G: k^{-1}gk\in H\}$ and define two maps $p_1,p_2:E\rightarrow G$ given by $p_1(g,k)=g$, and $p_2(g,k)=k$. I have just shown that $p_2$ is surjective. Now I want to show that for $k\in G$ there is a bijection from $p_2^{-1}(\{k\})$ to the subgroup $kHk^{-1}$.
I know that I could find a function between these two sets and show that this function is bijective and also a morphism, but I thought if it also works as follows:
By definition $$p_2^{-1}(\{k\})=\{(g,k)\in G\times G:k^{-1}gk\in H\}=\{g\in G:k^{-1}gk\in H\}=\{g\in G:g\in kHk^{-1}\}=kHk^{-1}$$ Now I thought that we have immediately show that these sets are bijective, since they are equal, but I'm not sure if this works.
Could someone take a look at my solution?
Thanks for your help.