Exercise 3.2.20 (If $A \unlhd G$ and $B \leq G$ then $A \cap B \unlhd AB$)

Proof. By Exercise 3.1.24, since $A⊴G$,

Since $A$ is abelian, $A$ commute with the the elements of the subgroup $A\cap B$ of $A$, thus

Therefore the normalizer $N_G(A\cap B)$ contains $A$ and $B$:

Moreover, since $A⊴G$, $\mathit{𝐴𝐵}$ is a subgroup of $G$, and it is the smallest subgroup of $G$ which contains $A$ and $B$ (suppose that $A\le H\le G$ and $B\le H$; if $h\in \mathit{𝐴𝐵}$, then $h=\mathit{𝑎𝑏}$, where $a\in A$ and $b\in B$, then $h=\mathit{𝑎𝑏}\in H$, so $\mathit{𝐴𝐵}\le H$).

Hence (1) implies

Therefore $\mathit{𝐴𝐵}$ normalizes $A\cap B$:

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