since b^{3}has order 4, the only conjugacy class out of [1],[a],[b],[b^{2}] b^{3}could possibly belong to is [b].

so assume that b^{3}= xbx^{-1}. we can automatically rule out all powers of b, since b commutes with all powers of b.

note we can write any element of F20 as a^{k}b^{m}, for k = 0,1,2,3,4 and m = 0,1,2,3 because of the relation bab^{-1}= a^{2},

so that ba = a^{2}b.

and (a^{k}b^{m})b(a^{k}b^{m})^{-1}= a^{k}ba^{5-k},

so it suffices to prove that no conjugate of b by a power of a gives b^{3}.

now aba^{-1}= aba^{4}= a(ba)a^{3}= a(a^{2}b)a^{3}= a^{3}(ba)a^{2}=

a^{3}(a^{2}b)a^{2}= ba^{2}= (ba)a = (a^{2}b)a = a^{2}(ba) = a^{4}b.

continuing: a^{2}ba^{-2}= a^{2}ba^{3}= a(aba^{-1})a^{4}= a(a^{4}b)a^{4}

= ba^{4}= (a^{2}b)a^{3}= a^{4}ba^{2}= a^{6}ba = aba = a^{3}b.

and next: a^{3}ba^{-3}= a^{2}(aba^{-1})a^{3}= a^{2}(a^{4}b)a^{3}

= aba^{3}= a(ba)a^{2}= a(a^{2}b)a^{2}= a^{3}(ba)a = a^{5}ba = ba = a^{2}b.

and: a^{4}ba^{-4}= a^{4}(ba) = a^{4}(a^{2}b) = ab.

so [b] = {b,ab,a^{2}b,a^{3}b, a^{4}b}.

since b^{3}is not in this conjugacy class, [b^{3}] must be a 5th conjugacy class.