Transitive normality satisfies image condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., transitively normal subgroup) satisfying a subgroup metaproperty (i.e., image condition)
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Statement

Statement with symbols

Suppose $H$ is a transitively normal subgroup of a group $G$ . Suppose $\varphi :G\to K$ is a surjective homomorphism of groups. Then, $\varphi (H)$ is a transitively normal subgroup of $K$ .

Related facts

Similar facts about similar properties

Related facts about transitively normal subgroups

Transitive normality is not quotient-transitive

Facts used

Proof

Given: A group $G$ , a subgroup $H$ . A surjective homomorphism $\varphi :G\to K$ of groups. $M=\varphi (H)$ . $L$ is a normal subgroup of $M$ .

To prove: $L$ is normal in $K$ .

Proof:

$N=H\cap \varphi ^{-1}(L)$ is normal in $H$ and $\varphi (N)=L$ : Let $\varphi _{0}:H\to M$ be the restriction of $\varphi$ . Then, $\varphi _{0}$ is a surjective homomorphism by definition, and fact (1) yields that $\varphi _{0}^{-1}(L)$ is normal in $H$ . Further, $\varphi _{0}^{-1}(L)$ clearly surjects to $L$ , since $\varphi _{0}$ is surjective. But $\varphi _{0}^{-1}(L)=H\cap \varphi ^{-1}(L)$ by definition, so $N=H\cap \varphi ^{-1}(L)$ is normal in $H$ and $\varphi (N)=L$ .
$N$ is normal in $G$ : From the previous step, $N$ is normal in $H$ . By assumption, $H$ is transitively normal in $G$ . Thus, $N$ must be normal in $G$ .
$\varphi (N)=L$ is normal in </math>K</math>: This follows from fact (2).