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.

$\blacksquare$


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