Intersection with Normal Subgroup is Normal/Examples/Subset Product of Intersection with Intersection

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Theorem

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 $x_h \in H_1 \cap H_2$.

Note that as $x_h \in H_1 \cap H_2$ it follows that $x_h \in H_1$ and $x_h \in H_2$.


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:

\(\displaystyle x_h \paren {H_1 \cap N_2} \paren {H_2 \cap N_1} x_h^{-1}\) \(=\) \(\displaystyle x_h \paren {H_1 \cap N_2} {x_h}^{-1} x_h \paren {H_2 \cap N_1} x_h^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {H_1 \cap N_2} x_h \paren {H_2 \cap N_1} x_h^{-1}\) as $H_1 \cap N_2 \lhd H_1$
\(\displaystyle \) \(=\) \(\displaystyle \paren {H_1 \cap N_2} \paren {H_2 \cap N_1}\) as $H_2 \cap N_1 \lhd H_2$

$\blacksquare$

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