Subset Product of Normal Subgroups is Normal
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $N$ and $N'$ be normal subgroups of $G$.
Then $N N'$ is also a normal subgroup of $G$.
Proof
From Subset Product with Normal Subgroup is Subgroup‎, we already have that $N N'$ is a subgroup of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From Subgroup is Normal iff Contains Conjugate Elements:
- $g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
- $\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$
So $N N'$ is normal.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \zeta$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $9$