An Exercise from Rotman's Theory of Groups (2.31 in my edition)?

suppose |G|<|S|+|T| and G is not equal to ST. let g in G such that g is not equal to any product st in ST. then s^(-1)g is not in T for any s in S: if s^(-1)g=t, then g=st by multiplying both sides on the left by s. let S^(-1) denote the set {s^(-1) | s in S}. so S^(-1)g intersect T is empty. the order of S^(-1)g is |S|. to see this, note that if s^(-1)g=s'^(-1)g for some s and s' in S, then, multiplying both sides on the right by g^(-1), we have s^(-1)=s'^(-1). this means that s=s' which means that |S^(-1)g| is greater than or equal to |S|. and clearly S^(-1)g is of order less than or equal to |S|, so it is of order |S|. hence we have found two disjoint subsets of G, one of order |S| and one of order |T|. so the order of G must be at least |S|+|T|. this contradicts our hypothesis that |G|<|S|+|T|.