Intersection with Normal Subgroup is Normal/Examples/Subset Product of Intersection with Intersection
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Example of Use of Intersection with Normal Subgroup is Normal
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let:
- $N_1 \lhd H_1$
- $N_2 \lhd H_2$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
- $\paren {H_1 \cap N_2} \paren {H_2 \cap N_1} \lhd \paren {H_1 \cap H_2}$
Proof
Consider arbitrary $x_h \in H_1 \cap H_2$.
By definition of set intersection, it follows that:
- $x_h \in H_1 \land x_h \in H_2$
We aim to show that:
- $x_h \paren {H_1 \cap N_2} \paren {H_2 \cap N_1} x_h^{-1} = \paren {H_1 \cap N_2} \paren {H_2 \cap N_1}$
thus demonstrating:
- $\paren {H_1 \cap N_2} \paren {H_2 \cap N_1} \lhd \paren {H_1 \cap H_2}$
by the Normal Subgroup Test.
First note that from Intersection with Normal Subgroup is Normal:
- $H_1 \cap N_2 \lhd H_1$
- $H_2 \cap N_1 \lhd H_2$
We have:
\(\ds x_h \paren {H_1 \cap N_2} \paren {H_2 \cap N_1} x_h^{-1}\) | \(=\) | \(\ds x_h \circ \paren {H_1 \cap N_2} {x_h}^{-1} \circ x_h \paren {H_2 \cap N_1} x_h^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x_h \circ \paren {H_1 \cap N_2} \paren{ {x_h}^{-1} \circ x_h } \paren {H_2 \cap N_1} x_h^{-1}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds x_h \paren {H_1 \cap N_2} e \paren {H_2 \cap N_1} x_h^{-1}\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x_h \paren {H_1 \cap N_2} \paren {H_2 \cap N_1} x_h^{-1}\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {H_1 \cap N_2} x_h \paren {H_2 \cap N_1} x_h^{-1}\) | as $H_1 \cap N_2 \lhd H_1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {H_1 \cap N_2} \paren {H_2 \cap N_1}\) | as $H_2 \cap N_1 \lhd H_2$ |
$\blacksquare$
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Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \nu$