Subset Products of Normal Subgroup with Normal Subgroup of Subgroup

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Theorem

Let $G$ be a group.

Let:

$(1): \quad H$ be a subgroup of $G$
$(2): \quad K$ be a normal subgroup of $H$
$(3): \quad N$ be a normal subgroup of $G$


Then:

$N K \lhd N H$

where:

$N K$ and $N H$ denote subset product
$\lhd$ denotes the relation of being a normal subgroup.


Proof

Consider arbitrary $x_n \in N, x_h \in H$.

Thus:

$x_n x_h \in N H$

We aim to show that:

$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$

thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.

We have:

\(\ds x_n x_h N K \paren {x_n x_h}^{-1}\) \(=\) \(\ds x_n x_h N K {x_h}^{-1} {x_n}^{-1}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds x_n N x_h K {x_h}^{-1} {x_n}^{-1}\) as $N \lhd G$ and $H \le G$
\(\ds \) \(=\) \(\ds x_n N K {x_n}^{-1}\) as $K \lhd H$
\(\ds \) \(=\) \(\ds N K {x_n}^{-1}\)
\(\ds \) \(=\) \(\ds K N {x_n}^{-1}\) as $N \lhd G$ and $K \le G$
\(\ds \) \(=\) \(\ds K N\)
\(\ds \) \(=\) \(\ds N K\) as $N \lhd G$ and $K \le G$

$\blacksquare$


Sources

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