Normal Subgroup of Subset Product of Subgroups
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Theorem
Let $G$ be a group whose identity is $e$.
Let:
- $H$ be a subgroup of $G$
- $N$ be a normal subgroup of $G$.
Then:
- $N \lhd N H$
where:
- $\lhd$ denotes normal subgroup
- $N H$ denotes subset product.
Proof
From Subset Product with Normal Subgroup is Subgroup:
- $N H = H N$ is a subgroup of $G$.
By definition of subset product all elements of $H N$ can be written in the form:
- $h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $n_1 \in N$.
From Inverse of Group Product:
- $\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$
We have that:
- $n n_1 n \in N$
- $h, h^{-1} \in G$.
Then, since $N$ is a normal subgroup of $G$:
- $\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$
Thus:
- $N \lhd N H$
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \epsilon$