Understanding the proof and meaning of the Butterfly lemma (Zassenhaus) (Lang's Algebra, pp. 20--21) I would like to type out my understanding of the Butterfly (Zassenhaus) Lemma, using the notation from pp. 20--21 of Lang's Algebra. I have trouble understanding Lang's proof so this is a hybrid of my own work and other sources.
Butterfly lemma:



Proof:
First off, it is immediate that $U \cap V, U \cap v$, and $u \cap V$ are subgroups of the unnamed group within which we are working. It is also clear that the latter two are subgroups of the former, since they are subsets of $U \cap V$ and groups themselves. We want to show that the latter two are in fact normal subgroups of $U \cap V$.
To see that $u \cap V$ is normal in $U \cap V$, consider $x \in u \cap V$ and $z \in U \cap V$. Because $x \in u$ and $z \in U$, by normality we see that $zx z^{-1} \in u$. Because $x \in V$ and $z \in V$, by closure we see that $zxz^{-1} \in V$. This shows that $zxz^{-1} \in u \cap V$ which proves normality of $u \cap V$ in $U \cap V$. An entirely symmetrical argument works to show that $U \cap v$ is normal in $U \cap V$.
Since the product of normal subgroups is still normal, we are able to form the quotient group $\frac{U \cap V}{(u \cap V)(U \cap v)}$, which we will return to later.
Next we want to show that $u(U \cap v), u(U \cap V), (u \cap V)v$, and $(U \cap V)v$ are groups. Because $U \cap v$ is a subgroup of the normalizer of $u$, i.e. $U$ itself, it follows that $u(U \cap v)$ is a group. The same argument works to show that $u(U \cap V)$ is a group, and a symmetrical argument works to show that the latter two are groups. (I think these product groups could be written in either order, in terms of the set multiplication, but I haven't checked this. I'm just following Lang's notation here.)
Now we want to show that $u(U \cap v)$ is normal in $u(U \cap V)$ and $(u \cap V)v$ is normal in $(U \cap V)v$. (It is clear that the first and third are subgroups of the second and fourth respectively, since they are clearly subsets and are groups themselves.) Consider the function $f \colon u(U \cap V) \to \frac{U \cap V}{(u \cap V)(U \cap v)}$ given by $ab \mapsto b(U \cap v)(u \cap V)$, where $a \in u$ and $b \in U \cap v$.
This is a well-defined function because if we take $a' \in u$ and $b' \in U \cap V$ such that $a'b' = ab$, then $a^{-1}a' = bb'^{-1}$ so by the left-hand side, $a^{-1}a' \in u$, and by the right-hand side, $bb'^{-1} \in U \cap V$, so $a^{-1}a' = bb'^{-1}$ must be in $u \cap (U \cap V) = u \cap V \subseteq (u \cap V)( U \cap v)$. This means that the inverse of $f(a'b')$ is the inverse of $f(ab)$, which implies that $f(a'b') = f(ab)$.
Furthermore $f$ is a homomorphism, which can be seen as follows. Consider $a, \alpha \in u$ and $b, \beta \in U \cap V$. We want to show that $f(ab \alpha \beta) = f(ab) f(\alpha \beta)$. By normality $b \alpha b^{-1} = \alpha' \in u$, so $\alpha = b^{-1} \alpha' b$, which means that $ab \alpha \beta = ab b^{-1} \alpha' b \beta = a \alpha' b \beta$. Therefore $f(ab \alpha \beta) = f(a \alpha' b \beta) = b \beta (u \cap V)(U \cap v) = f(ab) f(\alpha \beta)$.
The homomorphism $f$ is surjective, because for any $x \in U \cap V$, $f(ex) = x(u \cap V)(U \cap v)$. As for the kernel of $f$, we are looking for $ab \in u(U \cap V)$ such that $f(ab) = (u \cap V)(U \cap v)$. Clearly any $a \in u$ will suffice, and we need $b \in U \cap V$ by definition, but also $b \in (u \cap V)(U \cap v)$, and the intersection of those two is just $(u \cap V)(U \cap v)$. We can therefore write $b = xy$, where $x$ is an element of $u$ and $y$ is an element of $U \cap v$, which gives us $ab = axy = (ax)y \in u(U \cap v)$. This shows that $\ker(f) \subseteq u(U \cap v)$. On the other hand, if $cd \in u(U \cap v)$, then because $U \cap v \subset (u \cap V)(U \cap v)$, we see that $f(cd) = (u \cap V)(U \cap v)$ which shows that $u(U \cap v) \subseteq \ker(f)$. Thus $\ker(f) = u(U \cap v)$, so $u(U \cap v)$ is normal in $u(U \cap V)$.
By one of the isomorphism theorems, this establishes an isomorphism $\frac{u(U \cap V)}{u(U \cap v)} \cong \frac{U \cap V}{(u \cap V)(U \cap v)}$. By a symmetrical argument we conclude that $\frac{u(U \cap V)}{u(U \cap v)} \cong \frac{(U \cap V)v}{(u \cap V)v}$, as desired.
Comments:
Corrections or comments regarding my proof are appreciated. If anyone can explain Lang's proof in a way that I understand, I'd appreciate that too, because I don't get it. Lastly, if anyone can give me a decent intuition or main takeaway of this lemma, that would be great, because I'm not going to remember the details of this proof.
Lang's proof:





 A: Disclaimer: I was also trying to understand the intuition behind the Butterfly lemma, and I came across this post with no answer. So I decided to write an answer to help other future self-learners.
This answer is based on https://math.berkeley.edu/~gbergman/.C.to.L/ asserting some of my understandings
If $G$ is a group, and $U/u$, $V/v$ are homomorphic images of subgroups of $G$, meaning that there are two surjective homomorphisms(Results of Fundamental theorem on homomorphisms): one is from $U$ to $U/u$, and the other one is from $V$ to $V/v$.
After the definitions of the objects, one would like to
describe the extent to which one can "relate" part of the structure of $U/u$ and part of the structure of $V/v$, based on their common origin in $G$.
To find the common "region", one can take the subgroup of $U/u$ consisting of all elements which are also images of elements of $V$, which is written this as $u(U \cap V)/u$. Similarly, the analogous subgroup of $V/v$ is $(U \cap V)v/v$.
To get a common homomorphic image of these, we must divide each by the subgroup of those elements that are annihilated in the construction of the other. Namely, use $u(U \cap v)$ instead of $u$ in the denumerator and $(u \cap V)v$ instead of $v$ in the respective denumerator.
Now, the Butterfly Lemma says that after making these adjustments, we do get isomorphic groups, the "common heritage" of $U/u$ and $V/v$.
Remark: Group-theorists call a factor-group of a subgroup of a group $G$ a subfactor of $G$. Thus, given two subfactors $U/u$ and $V/v$ of a group $G$, the Butterfly Lemma characterizes their largest natural
"common subfactor".
