Prove these subgroups are equal I'm trying to solve this question which I found very complex and difficult due to the amount of details.

Let $H, K, N$ be subgroups of a group G such that $H\leq K$, $H\cap N
=K\cap N$, and $HN=KN$. Show that $H=K$.

I'm trying to prove it using this:

and this:

Am I in the right way? I really need a hand here, there are so many data in this question and I'm a little confused.
 A: Your assumption is $H \le K$, and your conclusion is $H = K$.
We already know $H \subseteq K$, let's show the opposite inclusion $K \subseteq H$.
Let $a \in K$. Then $a = a \cdot 1 \in KN = HN$, so $a = b n$ for some $b \in H$, $n \in N$. Thus $b^{-1} a = n \in K \cap N = H \cap N \subseteq H$, so $b^{-1} a \in H$, which means $a \in H$.


Note that this is related to Dedekind's modular identity, which states that if $N$ is a normal subgroup of a group $G$, and $H \le K \le G$, then
  $$K \cap N H = (K \cap N) H.\tag{Ded}$$
  This is a vestigial form of the distributive property, and lies at the core of the current argument.
In fact, using (Ded) and the hypotheses in OP one has
  $$
K = K \cap NK = K \cap N H = (K \cap N) H = (H \cap N) H = H.
$$

The proof of (Ded) goes exactly like the argument above. Clearly the inclusion $\supseteq$ holds. Conversely, let $a = n b \in K \cap N H$, with $a \in K$, $n \in N$, $b \in H$. Thus $a b^{-1} = n \in K \cap N$, so $a = (a b^{-1}) b \in (K \cap N) H$.
A: Proposition: If $H \leq K$, $H\cap N = K \cap N$, and $HN=KN$, then $H=K$.
Hint: Write an element of $K \leq KN =HN$ as an element of $HN$. A couple of arithmetic steps shows this element is already in $H$.

 Proof: Write $k \in K \leq KN = HN$ as $k=hn$, then $n=h^{-1}k \in K \cap N = H \cap N$, so both $h$ and $h^{-1}k$ are in $H$, so their product $k$ is also in $H$ and $K \leq H$. $\square$

A: We have


*

*$H\lt K$

*$H\cap N =K\cap N$

*$HN=KN$


The second implies that:


*

*$\forall h = n,\,\,\, \exists k, \,\,\,h = n = k$

*$\forall k = n,\,\,\, \exists h, \,\,\,h = n = k$


The third implies


*

*$\forall hn,\,\,\, \exists k\bar n,\,\,\, hn = k \bar n$

*$\forall k \bar n,\,\,\, \exists h n,\,\,\, hn = k \bar n$


and putting $n,\bar n = 1$ gives


*

*$\forall h,\,\,\, \exists k n,\,\,\, h = k n$

*$\forall k,\,\,\, \exists h n,\,\,\, k = h n$


since $H < K$ the first is trivial so we will take the second as the starting point.

For all $k$ there exists $h$, $n$ such that $k = h n$ or $h^{-1} k = n$ LHS is $\in K$ and RHS is $\in N$ so exists $\bar h$ such that $k = h n = h \bar h$.
This shows that every element of $K$ may be written in elements from $H$.
