I am working on excercise 1.9.19 of Howie's “Fundamentals of Semigroup Theory”:

Let I, J be ideals of a semigroup S. Show that I$\cap$J and I$\cup$J are ideals of S.

I am really struggling with proving this part of the question.

From the books properties of an ideal we have:

A non-empty subset I of S is called a left ideal if SI$\subseteq$I.

The right ideal is defined dually and an ideal has both.

I am wondering if it is as simple as:

(I $\cap$ J)S $\subseteq$ IS $\cap$ JS $\subseteq$ I $\cap$ J

and the equivalent for the union?

We then need to go on to prove:

Show also that $$(I\cup J)/J \cong I/(I\cap J)$$

For this I have seen similar proofs that start with a homomorphism from I to (I$\cup$J)/J and use properties of the image and kernal to arrive at an isomorphism but this also uses a theorem which I can't find within this book and would rather prove it using what I have.

  • $\begingroup$ So wha are the properties of an ideal and which of these are you struggling to show? $\endgroup$ Oct 30, 2018 at 13:47

1 Answer 1


Hint. If $I$ is an ideal of a semigroup $S$, then the semigroup $S/I$ can be represented as $(S \setminus I) \cup \{0\}$ where all products not falling in $S \setminus I$ are zero. Now just apply this construction to $(I\cup J)/J$ and to $I/(I\cap J)$ and see what's happen.


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