$[G:H]=\sum[K:(K\cap gHg^{-1})]$

Question

Let $G\geq H,K$ such that $[G:H]<\infty$ prove $[G:H]=\sum[K:(K\cap gHg^{-1})]$

my attempt
I've tried to make a group action of G on the coesets of H with conjugation and to try understand the orbits/stabilizers and I tried to figure out how the right index behaves so it might lead to me to know when the cosets have trivial intersection

Answer 1

Consider the coset space $G/H$, the subgroup $K$ should act on $G/H$ via left multiplication (not conjugation!) $k \cdot gH = kgH$. We can partition $G/H$ using the orbits of $K$ so, for a chosen set of representatives $\{g_i\}$ from each orbit we have: $$G/H = \bigsqcup_i {\rm Orb}(g_iH) = \bigsqcup_i Kg_iH. $$

Note that set $\{g_i\}$ is a set of representatives for the double coset space $K\backslash G/H$. In fact each $KgH$ corresponds to a unique orbit for the action of $K$ on $G/H$ (namely the orbit of the coset $gH$).

Now, the orbit-stabilizer theorem tells us that for any $gH \in G/H$ we have a bijection ${\rm Orb}(gH) \cong K/{\rm Stab}(gH)$. Who are the elements of these stabilizer subgroups? An element $k \in {\rm Stab}(gH)$ is an element of $K$ such that $kgH = gH$, in other words such that $g^{-1}kg \in H$ or $k \in gHg^{-1} = \ ^gH$.

This proves ${\rm Stab}(gH) \leq K \cap \ ^gH$ and we can easily prove the reverse inclusion, hence ${\rm Stab}(gH) = K \cap \ ^gH$.

Finally: $$[G: H] = |G/H| = \sum_i |K/{\rm Stab}(g_iH)| = \sum_i [K : {\rm Stab}(g_iH)] = \sum_i [K : K \cap \ ^gH].$$

Answer 2

The action is $K\curvearrowright G/H$ by left multiplication. The orbits correspond to double cosets $KgH$ (every orbit is a collection of cosets of $H$ whose union is a double coset). For every orbit, we can pick a particular representative $gH$, and then the size of the orbit equals $[K:\mathrm{Stab}(gH)]$ by the orbit-stabilizer theorem. The stabilizer $\mathrm{Stab}(gH)$ is $K\cap gHg^{-1}$, so the result follows.