Set of Subgroups of Abelian Group form Subsemigroup of Power Structure
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Theorem
Let $\struct {G, \circ}$ be an abelian group.
Let $\struct {\powerset G, \circ_\PP}$ denote the power structure of $\struct {G, \circ}$.
Let $\SS$ be the set of all subgroups of $G$.
Then $\struct {\SS, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
Proof
From Power Structure of Semigroup is Semigroup:
- $\struct {\powerset S, \circ_\PP}$ is a semigroup.
Let $A$ and $B$ be arbitrary subgroups of $G$.
We need to show that $A \circ_\PP B$ is also a subgroup of $G$.
Let $x$ and $y$ be arbitrary elements of $A \circ_\PP B$.
Then:
\(\ds \exists a_x \in A, b_x \in B: \, \) | \(\ds x\) | \(=\) | \(\ds a_x \circ b_x\) | |||||||||||
\(\ds \exists a_y \in A, b_y \in B: \, \) | \(\ds y\) | \(=\) | \(\ds a_y \circ b_y\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y^{-1}\) | \(=\) | \(\ds \paren {a_x \circ b_x} \circ \paren {a_y \circ b_y}^{-1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_x \circ b_x} \circ \paren { {b_y}^{-1} \circ {a_y}^{-1} }\) | Inverse of Group Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_x \circ {a_y}^{-1} } \circ \paren {b_x \circ {b_y}^{-1} }\) | Definition of Abelian Group: $\circ$ is commutative | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists a \in A, b \in B: \, \) | \(\ds x \circ y^{-1}\) | \(=\) | \(\ds a \circ b\) | where $a = a_x \circ {a_y}^{-1}$ and $b = b_x \circ {b_y}^{-1}$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y^{-1}\) | \(\in\) | \(\ds A \circ_\PP B\) |
Hence the result from the One-Step Subgroup Test.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.9$