Subset Product with Normal Subgroup is Subgroup
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Theorem
Let $G$ be a group whose identity is $e$.
Let:
- $(1): \quad H$ be a subgroup of $G$
- $(2): \quad N$ be a normal subgroup of $G$.
Let $H N$ denote subset product.
Then $H N$ and $N H$ are both subgroups of $G$.
Proof
It is clear that $e \in N H$, so $N H \ne \O$.
Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$.
Then:
\(\ds \paren {n_1 h_1} \paren {n_2 h_2}\) | \(=\) | \(\ds n_1 \paren {h_1 n_2 h_1^{-1} h_1} h_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n_1 \paren {h_1 n_2 h_1^{-1} } \paren {h_1 h_2}\) |
Since $N$ is normal in $G$:
- $\exists n \in N: n = h_1 n_2 h_1^{-1}$
Thus:
\(\ds \paren {n_1 h_1} \paren {n_2 h_2}\) | \(=\) | \(\ds \paren {n_1 n} \paren {h_1 h_2}\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds N H\) |
Also:
\(\ds \paren {n_1 h_1}^{-1}\) | \(=\) | \(\ds h_1^{-1} n_1^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {h_1^{-1} n_1^{-1} h_1} h_1^{-1}\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds N H\) |
so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$.
The fact that $N H = H N$ follows from Subset Product of Subgroups.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.16$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Problem $\text{HH}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \epsilon$
- 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 5$: Groups $\text{I}$: Subgroups
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $9$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.10$