Nicolas Boubarki, Algebra I, Chapter 1, § 2, Ex. 12

Question

Nicolas Boubarki, Algebra I, Chapter 1, § 2, Ex. 12:

($E$ is a Semigroup with associative law (represented multiplicatively), $\gamma_a(x)=ax$.)

Under a multiplicative law on $E$, let $ a \in E $ be such that $\gamma_a $ is surjective.

(a) Show that, if there exists $u$ such that $ua=a$, then $ux=x$ for all $x\in E$.

(b) For an element $b\in E$ to be such that $ba$ is left cancellable, it is necessary and sufficient that $\gamma_a$ be surjective and that $b$ be left cancellable.

For those interested in part (a), simple proof is that for every $x\in E$ there exists $x^\prime \in E$ such that $ax^\prime=x$, consequently $ua=a \Rightarrow uax^\prime=ax^\prime \Rightarrow ux=x$.

In (b), surjectivity of $\gamma_a$ and left cancellability of $b$ is required. However, I am concerned with "sufficiency" portion of part (b). When $E$ is infinite set there can always be a surjective function $\gamma_a$ which need not be injective, and left translation by $b$ is cancellable, however $ba$ need not be left cancellable.

Could you please post the full text of the exercise? In particular, what are $E$ and $\gamma_a$? — A.P., Apr 15 '13 at 19:02

score 3 · Answer 1 · answered Apr 15 '13 at 20:58

3

I have a Russian translation of Bourbaki. In it Ex.12 looks as follows:

"For $\gamma_{ba}$ to be an one-one mapping of $E$ into $E$, it is necessary and sufficient that $\gamma_{a}$ be an one-one mapping of $E$ onto $E$ and $\gamma_{b}$ be an one-one mapping of $E$ into $E$."

So I guess that there is a misprint in English translation. I wonder how it looks in the French original?

answered Apr 15 '13 at 20:58

Boris Novikov

17,470

It looks like this: "Pour qu'un élément $b\in E$ soit tel que $ba$ soit simplifiable à gauche, il faut et il suffit que $\gamma_a$ soit surjective et que $b$ soit simplifiable à gauche." – A.P. Apr 15 '13 at 21:42
@A.P.: So it is equal to English text? Well, then there is a misprint in the translation from Russian into French :-) – Boris Novikov Apr 15 '13 at 21:52
@BorisNovikov - Thanks a lot. – nature1729 Apr 16 '13 at 04:42
Well, note that $\gamma_c$ is injective iff $c$ is left cancellable, for some fixed $c\in E$. Thus the two formulations are equivalent. – A.P. Apr 16 '13 at 06:55
@A.P.: Which two? F and R, or F and E? – Boris Novikov Apr 16 '13 at 06:59
The French and English ones are literal translations of each other. The Russian one is an equivalent formulation, given in terms of the maps $\gamma$. – A.P. Apr 16 '13 at 07:06
@A.P.:No, in Russian formulation $\gamma_a$ is 1-1. – Boris Novikov Apr 16 '13 at 07:08
Actually, every map in the Russian formulation is 1-1, so I though it was some language oddity. Maybe not, after all... – A.P. Apr 16 '13 at 07:10
@A.P.:I think the Russian translator corrected the misprint. – Boris Novikov Apr 16 '13 at 07:27
So, $\gamma_b$ being an one-one mapping of $E$ onto $E$ should mean that it is bijective? I am more worried about semantics. – nature1729 Apr 16 '13 at 07:53
@Reetesh Mukul:Yes. – Boris Novikov Apr 16 '13 at 07:55
Summarizing necessity part, when $ba$ is cancellable, $\gamma_a$ must be bijective.
In Ex.13, Bourbaki is referring to this bijectivity property.
– nature1729 Apr 16 '13 at 08:08
@Reetesh Mukul: Take into account that in Ex.12 before (a) the next is written: "Let $a$ be such that $\gamma_a$ is surjective". – Boris Novikov Apr 16 '13 at 08:52

Nicolas Boubarki, Algebra I, Chapter 1, § 2, Ex. 12

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