How to show that S/J is a chain where S is a semigroup with no proper generating sets I'm Struggling with exercise 4.7.17 of Howies Fundamentals of Semigroup theory

Let $S$ be a non-trivial semigroup with the property that no proper subset of $S$ generates $S$. Show that each J−class of $S$ is a left zero semigroup or a right zero semigroup, and that the semilattice $S/J$ is a chain.

I really just can't seem to see how the given information helps to reach the conclusion, any help appreciated!
 A: Suppose $S$ is non-trivial and no proper subset of $S$ generates $S$ and suppose that $J$ is a $\mathscr{J}$-class of $S$. If there exist $x,y\in J$ such that $xy\not\in \{x,y\}$, then $S\setminus\{xy\}$ generates $S$, which contradicts your assumption. Hence for all $x,y\in J$ either $xy=x$ or $xy=y$. In particular, $x ^ 2 = x$ for all $x\in J$. So, $J$ is a subsemigroup of $S$, it is simple (it's a $\mathscr{J}$-class!) and every element is an idempotent. It follows that $J$ is isomorphic to a rectangular band $I\times\Lambda$ for some sets $I$ and $\Lambda$. 
You can prove that if $|I| > 1$ and $\Lambda| > 1$, then $I\times \Lambda$ (and so $J$ and $S$) is generated by a proper subset, which contradicts your assumption. Hence $J$ is either a left or right zero semigroup. 
If there exist $\mathscr{J}$-classes $J_1$ and $J_2$ that are incomparable in $S/\mathscr{J}$, and $x\in J_1$ and $y\in J_2$, then $xy\not\in J_1\cup J_2$. Again this means that $S\setminus \{xy\}$ is a generating set for $S$, which is a contradiction. It follows that $S/\mathscr{J}$ is a chain. 
