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 $\struct {G, \circ}$.
 * 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.

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:
 * $\paren {a \circ a'}^{-1} \in E$

Thus we have:

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:
 * $x = a \circ \paren {a^{-1} \circ x} \in a \circ E$

This shows that:
 * $A \subseteq a \circ E$

Thus, by definition of set equality:
 * $A = a \circ E$

Hence the result.