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 have just shown that $p_2$ is surjective, that there is a bijection from $p_2^{-1}(k)$ to $kHk^{-1}$ and that $\text{card}(E)=\text{card}(H)\cdot \text{card}(G)$. Now I have also shown that there is a bijection between $p_1^{-1}$ and $F_g=\{k\in G\mid g(kH)=kH\}=\{k\in G\mid kH\in (G/H)^g\}$. I also know that $\text{card}(F_g)=\text{card}((G/H)^g)\cdot \text{card}(H)$. I only need to show that $$\text{card}(E)=\sum_{g\in G} \text{card}((G/H)^g)\cdot \text{card}(H).$$
I somehow don't see the point. Could someone help me here? Thanks a lot!
Well, $$ \operatorname{card}(E) = \sum_{g \in G} \operatorname{card}(p_1^{-1}(g)) = \sum_{g \in G} \operatorname{card}(F_g) = \sum_{g \in G} \operatorname{card}((G / H)^g) \cdot \operatorname{card}(H) $$ by the work which you have already done. Presumably you did the same thing in order to show that $\operatorname{card}(E) = \operatorname{card}(G) \cdot \operatorname{card}(H)$.