Normality satisfies inverse image condition

Property-theoretic statement
The subgroup property of being normal satisfies the subgroup metaproperty called the inverse image condition: the inverse image of a normal subgroup, under a homomorphism, is normal.

Statement with symbols
Let $$\varphi:G \to H$$ be a homomorphism of groups, and $$N$$ be a normal subgroup of $$H$$. Then, $$\varphi^{-1}(N)$$ is a normal subgroup of $$G$$.

Related facts

 * Stronger than::Normality satisfies intermediate subgroup condition
 * Stronger than::Normality satisfies transfer condition

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

To prove: $$\varphi^{-1}(N)$$ is normal in $$G$$

Proof: Pick $$a \in \varphi^{-1}(N)$$ and $$g \in G$$. We need to show that $$gag^{-1} \in \varphi^{-1}(N)$$.

By the fact that $$\varphi$$ is a homomorphism:

$$\varphi(gag^{-1}) = \varphi(g)\varphi(a)\varphi(g)^{-1}$$

Since $$a \in \varphi^{-1}(N)$$, $$\varphi(a) \in N$$, and since $$N$$ is normal in $$H$$, the right side of the above equation is in $$N$$. Hence, $$\varphi(gag^{-1}) \in N$$, so $$gag^{-1} \in \varphi^{-1}(N)$$, as required.