Index in Subgroup

Theorem
Let $G$ be a group.

Let $H, K$ be subgroups of finite index of $G$.

Then:
 * $\index H {H \cap K} \le \index G K$

where $\index G K$ denotes the index of $K$ in $G$.

Equality happens $G = H K$.

Proof
We list out all the left cosets of $H \cap K$ in $H$:
 * $H / \paren {H \cap K} = \set {h_n \paren {H \cap K}: h_n \in H, n \in I}$

where $I$ is some finite indexing set.

For each pair $h_i, h_j \in H \subseteq G$, where $i \ne j$:
 * $h_i^{-1} h_j \notin H \cap K \quad$ Cosets are Equal iff Product with Inverse in Subgroup

Since $H$ is a subgroup, $h_i^{-1} h_j \in H$.

Thus:
 * $h_i^{-1} h_j \notin K$

By Cosets are Equal iff Product with Inverse in Subgroup, $h_i K$ and $h_j K$ are different left cosets.

Hence there must be at least $\index H {H \cap K}$ left cosets of $K$ in $G$, namely $h_n K$ for each $n \in I$.

It follows that $\index H {H \cap K} \le \index G K$.

Equality
Suppose equality holds.

Then the construction above gives all left cosets of $K$ in $G$.

For any $x \in G$, $x \in h_i K$ for some $h_i \in H$.

Hence $G \subseteq H K$.

The inclusion $H K \subseteq G$ is seen from.

Hence $G = H K$ by definition of set equality.

Now suppose $G = H K$.

To show equality we must show that the construction above gives all left cosets of $K$ in $G$.

Let $g K$ be a left coset of $K$ in $G$, where $x \in G$.

Since $G = H K$:
 * $\exists h \in H, k \in K: g = h k$

Then by Left Coset Equals Subgroup iff Element in Subgroup:
 * $g K = h \paren {k K} = h K$

Since $h \in H$:
 * $\exists i \in I: h \in h_i \paren {H \cap K}$

By Element in Left Coset iff Product with Inverse in Subgroup:
 * $h^{-1} h_i \in H \cap K \subseteq K$

Thus by Left Cosets are Equal iff Product with Inverse in Subgroup:
 * $h_i K = h K = g K$

Hence all left cosets of $K$ in $G$ are obtained in the construction.

This gives equality.