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.