Normality satisfies inverse image condition

This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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Statement

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

• Normality satisfies intermediate subgroup condition
• 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.