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