How to prove the butterfly lemma? Definitions:
If $K$ is a normal subgroup of a subgroup $H$ of $G$, then $H/K$ is referred to as a section of $G$.  
Two sections $H/K$, $H'/K'$ of the group $G$ are said incidents if any coset in $H/K$ has an intersection with one unique coset in $H'/K'$ that is not empty, and the converse.  
Let $L/M$ be one section of $G$ and $H$ a subgroup of $G$. We define the projection of $H$ upon $L/M$ to be the subset of $L/M$ that consists of cosets of $M$ in $L$ overlapping with $H$.

Exercise:  
i. Incident sections are isomorphic.  
ii. Let $N$ be one normal subgroup of $G$, $H$ one subgroup of $G$. Then $HN/N$ and $H/(H \cap N)$ are incident.   
iii. Conditions being as the definition iii., the projection of $H$ upon $L/M$ is the subgroup $(L \cap H)M/M$ of $L/M$. 
iv. The projection of $K'$ upon $H/K$ is a normal subgroup of the projection of $H'$ upon $H/K$. Then this quotient group is considered the projection of $H'/K'$ upon $H/K$. 
v. The projections, respectively, of $H/K$ upon $H'/K'$ and of $H'/K'$ upon $H/K$ are incident. 

The Final Result:  
Let $H$, $H'$ be subgroups of $G$, $K$ one normal subgroup of $H$, $K'$ that of $H'$. Then 
$ (H \cap {H'})K/(H\cap{K'})K $ is isomorphic to $(H\cap{H'})K'/(K\cap{H'})K'$.
It is taken from the book: Groups and representations, by JL Alperin, and Rowen B Bell.
Here $A\cap B$ is the intersection of them.
This was first proved by Zassebhaus, at the aureate age of 21, whereupon leaving the name of the lemma of Zassenhaus,  the fourth isomorphism theorem, or the butterfly lemma, owing to the shape of its inclusion diagram of involved subgroups.   
Question: I can without confronting many difficulties solve the first four exercises, while the last one still puzzles me. In effect, what confuses me is that, in order to consider two sections as incident, shall we not first be sure that they are sections with respect to the same group $G$? I cannot at this juncture even assume that the fifth exercise is true...
Thanks for any hint.
 A: Here is an example, since the comments are getting crowded:
Let $G = \{ 1, x, y, xy \}$ be a Klein 4-group.  The sections of G are precisely:


*

*$X/1 = \{ \{ 1 \}, \{ x \} \}$

*$Y/1 = \{ \{ 1 \}, \{ y \} \}$

*$Z/1 = \{ \{ 1 \}, \{ xy \} \}$

*$G/X = \{ \{ 1, x \}, \{ y, xy \} \}$

*$G/Y = \{ \{ 1, y \}, \{ x, xy \} \}$

*$G/Z = \{ \{ 1, xy \}, \{ x, y \} \}$

*$G/1 = \{ \{ 1 \}, \{ x \}, \{ y \}, \{ x, y \} \}$

*$1/1 = \{ \{ 1 \} \}$

*$X/X = \{ \{ 1, x \} \}$

*$Y/Y = \{ \{ 1, y \} \}$

*$Z/Z = \{ \{ 1, xy \} \}$

*$G/G = \{ \{ 1, x, y, xy \} \}$


The following (unordered) pairs of sections are incident:


*

*$(X/1,G/Y)$ since $\{1\} \cap \{1,y\}= \{1\}$, $\{1\} \cap \{x,xy\} = \varnothing$, $\{x\} \cap \{1,y\} = \varnothing$ and $\{x\} \cap \{x,xy\} = \{x\}$ (so that each coset in $X/1$ intersects exactly one coset in $G/Y$)

*$(X/1,G/Z)$

*$(Y/1,G/X)$

*$(Y/1,G/Z)$

*$(Z/1,G/X)$

*$(Z/1,G/Y)$

*all the $H/H$ are incident, since there is only one coset per section, and all the cosets contain the identity 1 of G
This is all (I believe).  For instance $G/X$ and $G/Y$ are not incident, since $\{1,x\}$ intersects both $\{1,y\}$ and $\{x,xy\}$.  Of course $G/X \to Z/1 \to G/Y$ shows the two sections are "connected" and so isomorphic, but they are not incident as in directly next to each other.
Similarly, $G/X \to Y/1 \to G/Z \to X/1$ shows that $G/X$ and $X/1$ are connected and isomorphic, but obviously not incident (sorry!) since $\{1,x\}$ intersects both $\{1\}$ and $\{x\}$ non-emptily.
A: Definition: If $H/K$ is a section of $G$ and $H'/K'$ is a section of $G$, then the projection of $H/K$ onto $H'/K'$ is the set of all cosets $hK'$ where h is in both H and H′
Lemma: The projection of a section of G onto a section of G is a section of G.
Proof: The projection of $H/K$ onto $H'/K'$ is precisely the section $((H\cap H')K')/K'$. $\square$
The original question only defines the projection of a subgroup, and obviously that subgroup needs to be a subgroup of G not $H/K$ for anything to make sense, so @awllower was understandably concerned.  Of course the projection of $H/K$ is the same as the projection of $H$, so perhaps the author felt no need to define it.
