Condition for Subgroup of Power Set of Group to be Quotient Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset G$, the power set of $G$.
Let $\struct {\LL, \circ_\PP}$ be a subgroup of the algebraic structure $\struct {\powerset G, \circ_\PP}$.
Then:
:there exists a subgroup $H$ of $G$
:and a normal subgroup $K$ of $H$
:such that $\struct {\LL, \circ_\PP}$ is the quotient group $H / K$
:
:the identity element of $\struct {\LL, \circ_\PP}$ is a subgroup of $\struct {G, \circ}$.
== Proof ==
From Power Structure of Group is Semigroup, we have that $\struct {\powerset G, \circ_\PP}$ is a semigroup.
=== Sufficient Condition ===
Let:
:there exist a subgroup $H$ of $G$
:and a normal subgroup $K$ of $H$
:such that $\struct {\LL, \circ_\PP}$ is the quotient group $H / K$.
Then the identity element of $\struct {\LL, \circ_\PP}$ is $e K = K$, which is a subgroup of $\struct {G, \circ}$.
=== Necessary Condition ===
Let the identity element of $\struct {\LL, \circ_\PP}$ be a subgroup of $\struct {G, \circ}$.
Denote this subgroup by $\struct {E, \circ}$.
It is to be shown that:
:there exists a subgroup $H$ of $G$
:and a normal subgroup $K$ of $H$
:such that $\struct {\LL, \circ_\PP}$ is the quotient group $H / K$.
We claim that:
:$H = \bigcup \LL$
and:
:$K = E$
To show that $\struct {\bigcup \LL, \circ}$ is a subgroup of $G$, we will use the Two-Step Subgroup Test.
We have that $\bigcup \LL$ is non-empty.
Let $a, b \in \bigcup \LL$ be arbitrary.
Then:
:$\exists A, B \in \LL: a \in A \land b \in B$
Since $\struct {\LL, \circ_\PP}$ is a group, by :
:$A \circ_\PP B \in \LL$
By definition of $\circ_\PP$:
:$a \circ b \in A \circ_\PP B$
Hence:
:$a \circ b \in \bigcup \LL$
showing that $\bigcup \LL$ is closed.
Since $\struct {\LL, \circ_\PP}$ is a group, by :
:$\exists A^{-1} \in \LL: A \circ_\PP A^{-1} = E$
Let $a' \in A^{-1}$ be arbitrary.
We have:
:$a \circ a' \in E$
Since $\struct {E, \circ}$ is a group, by again:
:$\paren {a \circ a'}^{-1} \in E$
By :
By Group Product Identity therefore Inverses:
:$a' \circ \paren {a \circ a'}^{-1}$ is an inverse of $a$.
Moreover:
:$a' \circ \paren {a \circ a'}^{-1} \in A^{-1} \circ_\PP E = A^{-1} \subseteq \bigcup \LL$
Therefore, every element of $\bigcup \LL$ has an inverse in $\bigcup \LL$.
By the Two-Step Subgroup Test:
:$\struct {\bigcup \LL, \circ}$ is a subgroup of $G$.
Now we will show that $E$ is a normal subgroup of $\bigcup \LL$.
We have:
:$a \circ E \circ a^{-1} \subseteq A \circ_\PP E \circ_\PP A^{-1} = A \circ_\PP A^{-1} = E$
:$a^{-1} \circ E \circ a \subseteq A^{-1} \circ_\PP E \circ_\PP A = A^{-1} \circ_\PP A = E$
and thus $E$ is by definition a normal subgroup.
Finally, we need to show the equality:
:$\paren {\bigcup \LL} / E = \LL$
We do this by showing that each set in $\LL$ is a (left) coset of $E$ in $\bigcup \LL$.
That is:
:$A = a \circ E$
We have:
:$a \circ E \subseteq A \circ_\PP E = A$
Let $x \in A$ be arbitrary.
We have:
:$a^{-1} \circ x \in A^{-1} \circ_\PP A = E$
Hence:
This shows that:
:$A \subseteq a \circ E$
Thus, by definition of set equality:
:$A = a \circ E$ 
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.12$
Category:Quotient Groups
Category:Subset Products
Category:Normal Subgroups