Transitive normality satisfies image condition

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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

Facts used

  1. Normality satisfies inverse image condition
  2. 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 </math>K</math>: This follows from fact (2).