Normality satisfies image condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., image condition)
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Statement

Property-theoretic statement

The subgroup property of being normal satisfies the image condition: the image of a normal subgroup under any surjective homomorphism is also normal.

Statement with symbols

Suppose $\varphi:G \to H$ is a surjective homomorphism of groups, and $N$ is a normal subgroup of $G$. Then, $\varphi(N)$ is normal in $H$.

Generalizations

This result is part of a more general result called the fourth isomorphism theorem (also called the lattice isomorphism theorem or correspondence theorem).

Proof

Given: $\varphi:G \to H$ is a surjective homomorphism of groups, and $N$ is a normal subgroup of $G$

To prove: $\varphi(N)$ is normal in $H$

Proof: Pick $a \in \varphi(N)$ and $b \in H$. We need to show that $bab^{-1} \in \varphi(N)$.

Since $a \in \varphi(N)$, there exists $g \in N$ such that $\varphi(g) = a$. Further, since $\varphi$ is surjective, there exists $h \in G$ such that $\varphi(h) = b$. Then:

$bab^{-1} = \varphi(h)\varphi(g)\varphi(g)^{-1} = \varphi(hgh^{-1})$

(where the second step uses the fact that $\varphi$ is a homomorphism).

Now, since $N$ is normal in $G$, $hgh^{-1} \in N$, and hence $\varphi(hgh^{-1}) \in \varphi(N)$, showing that $bab^{-1} \in N$.