Problem 15 from Herstein's book

Suppose that there exist an onto map $f:S\to U$. Since $T<U$, there exist an onto map $f_0:U\to T.$ So $f_0\circ f:S\to T$ is onto and, what is a contradiction since $S<T$.

Let $g:S \to U$ which is onto. Then $f_2 \circ g$ is onto $T$, a contradiction.