The intersection by an abelian normal subgroup is normal in the product

Let G be a group and A,B \leq G be subgroups such that A is abelian and normal in G. Prove that A \cap B is normal in AB.

First we prove a lemma.

Lemma: Let G be a group, let H,K,N \leq G be subgroups, and suppose N \vartriangleleft H. Then N \cap K \vartriangleleft H \cap K. Proof: Let a \in H \cap K. Then a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a. \square

Now A \cap B \vartriangleleft A because A is abelian and A \cap B \vartriangleleft B by the lemma. Now if x \in AB, x = ab for some a \in A and b \in B. Thus x(A \cap B) = ab(A \cap B) = a(A \cap B)b = (A \cap B)ab = (A \cap B)x.

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