Normality satisfies image condition

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 $$bab^{-1} \in N$$.