# 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 $K$.

## 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:

1. $N = H \cap \varphi^{-1}(L)$ 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 $\varphi(N) = L$.
2. $N$ 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 $G$.
3. $\varphi(N) = L$ is normal in [/itex]K[/itex]: This follows from fact (2).