2.21.

We prove the hint: Since A is abelian, if g is an element of A and a is an element of A ^ H, then a

^{g}=a, which implies (A ^ H)^{g}= A ^ H.Thus A is a subset of N_{G}(A ^ H). If h is in H and a is in A ^ H, then a^{h}is in H since both h and a are in H, and a^{h}is in A because A is normal in G. Then a^{h}is in A ^ H. Thus H is a subset of N_{G}(A ^ H). Since N_{G}(A ^ H) is a subgroup of G, we have that AH is a subgroup of N_{G}(A ^ H). Since AH=G, this implies N_{G}(A ^ H<)=G, so A ^ H is normal in G.