This is Exercise 2.6.1 of Howie's "Fundamentals of Semigroup Theory".

Definition 1: A semigroup $S$ is cancellative if for all $a, b, c$ in $S$, we have both $ca=cb\implies a=b$ and $ac=bc\implies a=b$.

A Typo:

Let $C$ be a cancellative semigroup without an identity. The first part of the question reads

Lemma 1: Show that there cannot be any pair of elements $e,a$ in $C$ for which $ea=a$ or for which $\color{red}{ae=e}$.

(Surely this is wrong since $a, e$ are arbitrary.)

Partial Proof of Lemma 1: Suppose otherwise and that $ea=a$. Then for any $b\in C$, $bea=ba$, so that $be=b;$ in particular, $ae=a$; then $e$ is an identity for an arbitrary $a\in C$, a contradiction. $\square$

Is this proof valid?

The Question:

Question 1: Deduce (from Lemma 1) that $\mathcal L=\mathcal R=\mathcal D= 1_C$, where $\mathcal L,\mathcal R,\mathcal D$ are Green's equivalences.

I'm stuck. I've written out the definitions.

Also, the next part of the question goes like this.

Let $$S=\left\{ \begin{pmatrix} a & 0 \\ b & 1 \end{pmatrix} : a, b\in\Bbb R, a, b>0\right\}$$. Then

Lemma 2: With respect to matrix multiplication, $S$ is a cancellative semigroup without an identity.

Proof: Clearly the identity matrix is not in $S$ and that $S$ is cancellative follows from the cancellative properties of $\Bbb R$. $\square$

Question 2: Show that $\mathcal J= S\times S$, and deduce that $\mathcal D$ is properly contained in $\mathcal J$.

Again, I'm stuck.

Please help :)

  • 1
    $\begingroup$ You proof of lemma 1 is a little confusing. Assuming to the contrary gives you that $ae=a$ and $ea=a,$ so you can deduce that $e$ is the identity immediately. I don't think that is precisely the contradiction you are hoping for. Since $aeb=ab,$ then $ae=a,$ and similarly $bea=ba$ gives $be=b,$ was what you had assumed when you assumed the contrary to begin with. $\endgroup$ May 10, 2017 at 13:58
  • $\begingroup$ Could you write down the definitions of $\mathcal{L,R,D,J}$? $\endgroup$ May 11, 2017 at 22:37
  • $\begingroup$ @DanielRobert-Nicoud I've added a link to the Wikipedia page defining them. $\endgroup$
    – Shaun
    May 11, 2017 at 22:43

1 Answer 1


Lemma 1 typo

Note that the two statements in Lemma 1 can be written "For all $e,a\in C$, $ea\neq a$" and "For all $e,a\in C$, $ae\neq e$". So I think you are right that the second one is a typo, not because it's false (it's true), but because it's the same as the first (after switching $e$ and $a$ which are dummy variables). It should read "For all $e,a\in C$, $ae\neq a$".

Lemma 1 proof

Your proof of Lemma 1 states that $a\in C$ was arbitrary, but this isn't true; $a$ was provided by your assumption that the statement was false. However you did show $be=b$ for arbitrary $b\in C$, so $e$ is a right identity. In particular $e^2=e$, so pulling the same trick, for any $b\in C$ we have $$ e^2b=eb\hspace{10mm}\Rightarrow\hspace{10mm}eb=b. $$ Thus $e$ is also a left identity. The proof of the second statement is analogous.

Question 1

If $a\mathcal L b$ then $a=xb$ and $b=ya$ for some $x,y\in S^1$. Thus $a=xya$. If $xy\in S$, this would contradict Lemma 1, so $xy=1$, implying $x=1$ and $y=1$ (by definition of $S^1$). Thus $a=b$, so $\mathcal L$ is trivial. Similarly $\mathcal R$ is trivial, so $\mathcal D$ is also trivial.

Question 2

We must show $x\mathcal Jy$ for arbitrary elements $$ x=\begin{pmatrix}a&0\\ b&1\end{pmatrix},\, y=\begin{pmatrix}c&0\\ d&1\end{pmatrix}. $$ We have $$\begin{eqnarray*} \begin{pmatrix}\frac{2c(a+b)}{ad}&0\\1&1\end{pmatrix} \begin{pmatrix}a&0\\b&1\end{pmatrix} \begin{pmatrix}\frac{d}{2(a+b)}&0\\\frac{d}2&1\end{pmatrix} \hspace{-60mm}\\ &=& \begin{pmatrix}\frac{2c(a+b)}{d}&0\\a+b&1\end{pmatrix} \begin{pmatrix}\frac{d}{2(a+b)}&0\\\frac{d}2&1\end{pmatrix}\\ &=& \begin{pmatrix}c&0\\d&1\end{pmatrix}. \end{eqnarray*}$$ Hence $y\in S^1xS^1$. Similarly $x\in S^1yS^1$, so $x\mathcal Jy$.


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