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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.20

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 $AB$.

Solution: 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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