### 50th Putnam 1989

**Problem 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?

**Solution**

Answer: yes.

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

50th Putnam 1989

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001