TITLE: Does the following identity hold?
QUESTION [5 upvotes]: I encountered a claim in the proof a a claim in a paper that seemed to imply/use the following:

Suppose we have two subgroups $K$ and $H$ of a larger group $G$.  Let $T$ be a translation of $K$, which is defined as a subset of $G$ such that every $g \in G$ can be uniquely written as $t k$, for some $t \in T$ and $k \in K$. For example, if $G = \mathbb{Z} / 12 \mathbb{Z}$ and $K = \{0, 2, 4, \cdots, 10\}$, then $T = \{0, 1\}$ is a possible translation of $K$.
Let us also for simplicity define the notation $t K = \{tk ~|~ k \in K\}$.
The claim is then the following: for any translation of $T$ and any subgroup $H \leq G$ which is neither a subset nor a superset of $K$, we have the following:
$$\sum_{t \in T} |tK \cap H|^2 = |H| \cdot |K \cap H|$$

Is this claim true? It does seem to work out for example with $G, T$ as given above and $H = \{0, 3, 6, 9\}$, but I can't seem to find a proof of this fact or even an intuition as to why this should be true. If it is not true, then are there any other conditions we can impose (maybe I have missed them in reading the paper) so that it is true?
EDIT: The paper in question can be found on ArXiv at this link. The question stems from a step in the proof of lemma $4$.

REPLY [2 votes]: This is basically the same argument as Benjamin's, but I dislike putting group elements in lists.
Let $T'=\{t\in T\mid tK\cap H=\varnothing\}$ as in the other answer.
Not only is $|tK\cap H|$ always either $0$ or $|K\cap H|$, but it's a coset of $K\cap H$. That is, if $s\in tK\cap H$ (i.e. the intersection is nonempty) then $s\in tK$ means $tK=sK$ and $s\in H$ means $H=sH$ so
$$ tK\cap H=sK\cap sH=s(K\cap H). $$
This yields a map $T'\to H/(K\cap H)$. It's actually a one-to-one correspondence.
Since the $tK$s (hence the $tK\cap H$s) are disjoint, this map is one-to-one, and it's also onto since $G/K$ is a partition so every coset of $H/(K\cap H)$ intersects some $tK$ (hence some $tK\cap H$).
This means we have
$$ \sum_{t\in T'} |tK\cap H|^2=\left|\frac{H}{K\cap H}\right||K\cap H|^2 = |H||K\cap H|. $$