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:
- $H = \bigcup \LL$
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 by hypothesis that $\bigcup \LL$ is non-empty.
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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 Group Axiom $\text G 0$: Closure:
- $A \circ_\PP B \in \LL$
- $a \circ b \in A \circ_\PP B$
Hence:
- $a \circ b \in \bigcup \LL$
showing that $\bigcup \LL$ is closed.
Since $\struct {\LL, \circ_\PP}$ is a group, by Group Axiom $\text G 3$: Existence of Inverse Element:
- $\exists A^{-1} \in \LL: A \circ_\PP A^{-1} = E$
Let $a' \in A^{-1}$ be arbitrary.
We have:
- $a \circ a' \in E$
Since $\struct {E, \circ}$ is a group, by Group Axiom $\text G 3$: Existence of Inverse Element again:
- $\paren {a \circ a'}^{-1} \in E$
By Group Axiom $\text G 2$: Existence of Identity Element:
\(\ds e\) | \(=\) | \(\ds \paren{a \circ a' } \circ \paren {a \circ a'}^{-1}\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {a' \circ \paren {a \circ a'}^{-1} }\) | Group Axiom $\text G 1$: Associativity |
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:
\(\ds x\) | \(=\) | \(\ds e \circ x\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ a^{-1} } \circ x\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {a^{-1} \circ x}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(\in\) | \(\ds a \circ E\) | as shown above |
This shows that:
- $A \subseteq a \circ E$
Thus, by definition of set equality:
- $A = a \circ E$
Hence the result.
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Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.12$