Transitive normality satisfies image condition

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 $$K$$.

Similar facts about similar properties

 * Central factor satisfies image condition
 * SCAB satisfies image condition
 * Direct factor satisfies image condition
 * Centrality satisfies image condition

Related facts about transitively normal subgroups

 * Transitive normality is not quotient-transitive

Facts used

 * 1) uses::Normality satisfies inverse image condition
 * 2) uses::Normality satisfies image condition

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 K : This follows from fact (2).