# Putnam: Understanding a proof that a cancellative semigroup $S$ s.t. for any $s\in S$ the set of powers of $s$ is finite, is a group.

$$B2.$$ Let $$S$$ be a non-empty set with a binary operation (written like multiplication) such that:

(1) it is associative;

(2) $$ab = ac$$ implies $$b = c$$;

(3) $$ba = ca$$ implies $$b = c$$;

(4) for each element, the set of its powers is finite.

Is $$S$$ necessarily a group?

I understand the solution given:

Let a be any element. We show that for some n > 1 we have an = a. The set of its powers is finite, so for some r > s we have ar = as. If s = 1, we are done. If not, put b = as-1, then b ar-s+1 = b a, so we may cancel to get an = a with n = r - s + 1 > 1. Now put e = an-1. Then we have ea = ae = a.

Now take any b. We have a(eb) = (ae)b = ab, and cancelling gives eb = b. Similarly, (be)a = b(ea) = ba, so be = b. Hence e is an identity.

Also a has an inverse. If n - 1 = 1, then a = e, so a is its own inverse. If n - 1 > 1, then an-2 is its inverse.

Now if b is any other element, we may use the same argument to find another identity f and an element c such that cb = bc = f. But we have e = ef = f, so the identity is unique and c is an inverse for b.

The problem I am having is that they don't mention anything about the closure of such set $$S$$ under the operation.

For example, take $$S = \{1,2,3\}$$ under multiplication. I wouldn't say this is a group because $$2\times3$$ is not in $$S$$ but it fills all the criteria of the problem. What am I misunderstanding about closure? Is it something so obvious they don't even need to mention it in the solution? Thanks.

• The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $\times:G^2\to G$ such that for any $a,b\in G$ there is a unique $c\in G$ and a unique $d\in G$ such that $a\times c=b=d\times a.$ – DanielWainfleet Jul 16 '17 at 1:53
• A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed. – fleablood Jul 16 '17 at 1:56
• @DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $a\in S$ we have $aS=S=Sa$. – Shaun Nov 28 '18 at 23:48

A "binary operation on a set $S$" is a map $S \times S \to S$. Multiplication on $S = \{1,2,3\}$ is a map $S \times S \to \{1,2,3,4,6,9\}$ not $\to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $\mu : S \times S \to S$ and I have some subset $T \subset S$. Then $\mu$ restricts to a map $T \times T \to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.