I have the following problem and I am a bit stuck.
Let $G$ be a finite group acting on $X$. We first consider the case that $X=G/H$ where $H$ is a subgroup of $G$. Here we consider the natural action $$G\times G/H \rightarrow G/ H,\,\,(g,xH)\mapsto gxH.$$ We additionally introduce the following quantity $$E=\{(g,k)\in G\times G\mid k^{-1}gk\in H\}.$$ Then we have the following two functions $p_1,p_2:E\rightarrow G$ with $p_1(g,k)=g$ and $p_2(g,k)=k$. I need to show that $p_2$ is surjective and for $k\in G$ there is a bijection between $p_2^{-1}(k)$ and the subgroup $kHk^{-1}$.
I would have done this as follows: (i) Let $k\in G$, we notice that $(1,k)\in E$, that is we have found an element in E such that $p_2(e,k)=k$. This in turn means that $p_2$ is surjective.
(ii) We now need to find a bijection $\phi:p_2^{-1}(k)\rightarrow kHk^{-1}$. Here I don't know how to choose this exactly, somehow I have to use the surjection to from $p_2^{-1}(k)$ G and then see that I find an injection to $kHk^{-1}$ or not?
Could someone help me here? Thanks a lot!
Consider the map $\phi(g) = g$. First, we note that for every $(g,k) \in p_2^{-1}(k)$, we have that $k^{-1}gk \in H$. Thus, we have that $$g = kk^{-1}gkk^{-1} \in kHk^{-1}.$$
Now clearly this map is injective. All we need to show is that it is surjective. For that pick $khk^{-1} \in kHk^{-1}$. Then clearly $k^{-1}khk^{-1}k = h \in H$. Thus, $(khk^{-1},k) \in p_2^{-1}(k)$. Thus, we have subjectivity. Hence it is a bijection.