2
$\begingroup$

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 :)

$\endgroup$
3
  • 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

1
+50
$\begingroup$

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 $x\mathcal Jy$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.