Normality satisfies inverse image condition

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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

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.

Facts about Normality satisfies inverse image conditionRDF feed

Page class	Fact +
Stronger than	Normality satisfies intermediate subgroup condition +, and Normality satisfies transfer condition +

Normality satisfies inverse image condition

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Statement

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