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

$\Box$

### 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:

$\ds H = \bigcup \LL$
$K = E$

To show that $\ds \struct {\bigcup \LL, \circ}$ is a subgroup of $G$, we will use the Two-Step Subgroup Test.

Pick any $\ds a, b \in \bigcup \LL$.

Then there exists $A, B \in \LL$ such that $a \in A$ and $b \in B$.

Since $\struct {\LL, \circ_\PP}$ is a group, by Group Axiom $\text G 0$: Closure:

$A \circ_\PP B \in \LL$.
$a \circ b \in A \circ_\PP B$

and hence $\ds a \circ b \in \bigcup \LL$, showing that $\ds \bigcup \LL$ is closed.

Since $\struct {\LL, \circ_\PP}$ is a group, by Group Axiom $\text G 3$: Existence of Inverse Element:

$A^{-1} \in \LL$ where $A \circ_\PP A^{-1} = E$

Pick any $a' \in A^{-1}$.

We have $a \circ a' \in E$.

Since $\struct {E, \circ}$ is a group, by Group Axiom $\text G 3$: Existence of Inverse Element again:

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

Thus we have:

$a \circ a' \circ \paren {a \circ a'}^{-1} = e$

So $a' \circ \paren {a \circ a'}^{-1}$ is an inverse of $a$.

Moreover:

$\ds a' \circ \paren {a \circ a'}^{-1} \in A^{-1} \circ_\PP E = A^{-1} \subseteq \bigcup \LL$

and therefore all elements of $\ds \bigcup \LL$ has an inverse in $\ds \bigcup \LL$.

By Two-Step Subgroup Test, $\ds \struct {\bigcup \LL, \circ}$ is a subgroup of $G$.

Now we will show that $E$ is a normal subgroup of $\ds \bigcup \LL$.

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 $\ds \paren {\bigcup \LL} / E = \LL$.

We can do this by showing that each set in $\LL$ is a (left) coset of $E$ in $\ds \bigcup \LL$, that is, $A = a \circ E$.

We have $a \circ E \subseteq A \circ_\PP E = A$.

Now pick any $x \in A$.

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 $A = a \circ E$ by definition of set equality, and hence the result.