Cardinality of $HgK$ with $H,K$ subgroup of a finite group $G$ with fundamental counting principle

Question

Let $G$ be a finite group. Let $g \in G$ and let $H,K$ be subgroup of $G$

I have to compute that $|HgK|=\frac{|H|\cdot |K|}{|H^g \cap K|}$ with the fundamental counting principle. I know that there are some other proof, but I need a proof that use the theorem before mentioned.

I think that the way is to consider the action $$K \to HgK$$ $$k \mapsto Hgk$$ So, $HgK=\cup_{k\in K} \mathcal{O}_{Hgk}$ (with $\mathcal{O}$, I mean the orbit). I can compute $$|\mathcal{O}_{Hg}|=|K:stab_K(Hg)|$$ since $stab_K(Hg)=stab_G(Hg)\cap K=H^g \cap K$, we obtain $$|\mathcal{O}_{Hg}|=|K:stab_K(Hg)|=|K:(H^g\cap K)|=\frac{|K|}{|H^g\cap K|}$$ Now, since I know that $|HgK|=\frac{|H|\cdot |K|}{|H^g \cap K|}$, I have to have that there are exactly $H$ different conjugation classes with same cardiality. It’s clear that they have same cardinality but I don’t understand why they are exactly |H|.

Answer 1

The intended answer is probably like this.

Define a map $f\colon H \times K \to G$ by $f(h, k) = hgk$.

The image of $f$ is, by definition, the set $HgK$.

Now using the "fundamental counting principle", you only need to show that, for any element $\tilde g \in HgK$, the inverse image $f^{-1}(\tilde g)$ has cardinality equal to $|H^g \cap K|$.

To prove that, write $\tilde g = \tilde h g \tilde k$ and define a map $\phi\colon H^g \cap K \to H \times K$ by $\phi(t) = (\tilde h gtg^{-1}, t^{-1}\tilde k)$. We show that $\phi$ is a bijection from $H^g \cap K$ to $f^{-1}(\tilde g)$.

$\phi$ is clearly injective; and it's easy to verify that $\phi$ has image in $f^{-1}(\tilde g)$. To show surjectivity, take any $(h, k)$ such that $hgk = \tilde g = \tilde h g \tilde k$ and rewrite it as $g^{-1}\tilde h^{-1}hg = \tilde k k^{-1}$. Since the left hand side belongs to $H^g$ and the right hand side belongs to $K$, this element (call it $t$) belongs to $H^g \cap K$ and we have $\phi(t) = (h, k)$.