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I don't understand how to get this:

A finite semigroup with cancellation laws is a group.

Thanks in advance for any help.

Shaun
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patang
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2 Answers2

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Let $S$ be a such semigroup. For each $a\in S$, we can find a positive integer $n(a)$ such that $a^{n(a)+1}=a$. If such $n(a)$ does not exist, then $a^1,a^2,a^3,\cdots$ are all different and is contradicting that $S$ is finite.

We will argue that, for each $a,b\in G$, $a^{n(a)}=b^{n(b)}$. Since

$$aa^{n(a)}b=a^{n(a)+1}b=ab=ab^{n(b)+1}=ab^{n(b)}b$$

so by (left and right) cancellation law, we get $a^{n(a)}=b^{n(b)}$.

Let define $e=a^{n(a)}$ then $ea=a$ and $ae=a$ for all $a\in S$. If we define $a^{-1}:=a^{n(a)-1}$ then you can check that $a^{-1}$ is an inverse of $a$.

Shaun
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Hanul Jeon
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  • This is a nice argument! I haven't seen this before although I'm concerned with semigroups for several years. It should be noted that cancellability is already applied in the first paragraph: indeed from the fact that the set ${a_1,a_2,a_3,…}$ is finite, in the first step you can only conclude that there are $k,n>0$, $n≠k$ such that $a^k=a^n$. Now you have to cancel until an equation of the form $a^ma=a$ remains. – user 59363 Nov 20 '14 at 12:23
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For every $a\in S$ consider the mappings $S\to S$, $x\mapsto ax$ and $x\mapsto xa$. The cancellation conditions imply that these mappings are injective. Since $S$ is finite they are also surjective. This means for any $a,b\in S$ the equations $ax=b$ and $xa=b$ admit solutions. This already guarantees that $S$ is a group. (The idea is the same as for the usual proof that each finite integral domain is a field.)

user 59363
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