Showing that $[H:(K\cap H)] \leq [G:K]$ For $H,K < G$ For this problem we want to find a function $\eta: H/(K \cap H) \to G/K$ and show that $\eta$ is well-defined and injective. For this problem we will need to use a result from an earlier part of this problem that says that:

$\forall g \in G$ $H \cap gK$ is either empty or equal to a coset of $K \cap H$ in $H$: $h(K \cap H), h \in H$


For this problem I believe I understand the reasoning behind what we are doing. In that,  $[H:(H\cap K)] \leq [G:K]$ is equivalent to saying that $\lvert H/(H \cap K) \rvert \leq \lvert G/K \rvert$ so by finding a function from $H/(H\cap K)$ to $G/K$ and showing it is well defined and injective then we are showing, by invoking the Pigeonhole Principle, that it cannot be the case that $$\lvert H/(H \cap K) \rvert \gt \lvert G/K \rvert$$ And hence the inequality in the problem statement must be true.
Elements of

*

*$H/(K\cap H)$ look like $h(K\cap H) = hK \cap hH = hK \cap H$ for any $h \in H$

*$G/K$ look like $gK$ for any $g \in G$
First off, I'm having a bit of trouble actually defining my function $\eta$ in that I'm not sure how to write the arbitrary element of $H/(K\cap H)$ that's being taken by $\eta$ to $G/K$. In the first part I found an $a \in H \cap gK$ such that $a = h = gk$ for $h \in H\ \text{and}\ k \in K$ so that $$a(K \cap H) = a(H \cap K) = aH \cap aK = H \cap aK$$ But I'm not quite seeing how this simplifies things.
Also, I've not seen many examples regarding 'well-definedness' but my understanding is that if we have $\eta(a) = a'$ and $\eta(b) = b'$ for $a,b, \in H/(K\cap H)$ then $\eta(ab) = a'b'$. Correct?
 A: Restating: 
Given two subgroups $H,K\leq G$, we want to show there exists a well-defined, injective function between the sets of cosets:
$$\eta: H/(H\cap K)\to G/K.$$

It can be shown that for any coset of the intersection in $H$, we have:
$$h(H\cap K) = hH\cap hK = H\cap hK.$$
And since $hK$ is a coset of $K$ in $G$, the map suggests itself as:
$$\eta:h(H\cap K)\mapsto hK.$$

(i) Suppose $h,h'\in H$ represent the same coset. Then $h(H\cap K) = h'(H\cap K)$. Multiplying by the inverse on the left yields:
$$H\cap K = (h^{-1}h')(H\cap K)$$
Hence $h^{-1}h'\in H\cap K$ and in particular, $h^{-1}h'\in K$. It follows then that $K = (h^{-1}h')K$. So that finally multiplying by $h$ again gives:
$$hK = h'K.$$
This gives well-definition of $\eta$. $\square$
Thanks to Izaak for the assist here!

(ii) Suppose for $h,h'\in H$ that:
$$hK = h'K.$$
Then intersecting both sides with $H$, we get the middle equality:
$$h(H\cap K) = hH\cap hK = H\cap hK \text{ }(=)\text{ } H\cap h'K = h'H\cap h'K = h'(H\cap K),$$
from which it follows that
$$h(H\cap K) = h'(H\cap K).$$
Hence $\eta$ is injective. $\blacksquare$
