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.