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

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

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

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

$ba{b}^{-1}=\phi (h)\phi (g)\phi (g{)}^{-1}=\phi (hg{h}^{-1})$

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

Now, since $N$ is normal in $G$, $hg{h}^{-1}\in N$, and hence $\phi (hg{h}^{-1})\in \phi (N)$, showing that $ba{b}^{-1}\in N$.