if aH = bH, this means that for h in H, ah = bh' (for some different element h' of H).

now if we look at a^2H, we have the element a^2h = a(ah) = a(bh').

from ah = bh', we can conclude that a = bh'h^-1, so a^2h = (bh'h^-1)(bh').

the problem is, b may not commute with elements of H, so we cannot say if bH = Hb.

if fact, if bH ≠ Hb, we can't take the second b in bh'h^-1bh' "across" the h'h^-1 part to get it with the first b to make it be b^2(stuff in H).

now, formalize this, by finding a small group G, with a subgroup H and an element g with gH ≠ Hg.

you'll want a non-abelian group. then find some other element g' of G, with g'H = gH, and (g')^2H ≠ g^2H.