# Prove that this finite set is a group

## Homework Statement

Let G be a nonempty finite set with an associative binary operation such that:
for all a,b,c in G
ab = ac => b = c
ba = ca => b = c
(left and right cancellation)
Prove that G is a group.

2. The attempt at a solution
Let a $\in$ G, the set <a> = {$a^k$ : k $\in$ N} is a finite closed subset of G. So, $\exists$($k_1$, $k_2$)$\in$N, such that:
$a^{k_1}=a^{k_2}$ using the cancellation property I found that $a=a.a^{k_2-k_1}$.
So,
$a^{k_2-k_1}$ is the identity but the problem is in this reasoning every cyclic subgroup will generate a different identity. And the identity is supposed to be unique. I don't know how to proceed now, any help would be appreciate it.
Thanks is advance,
NAta

## Answers and Replies

Deveno
Science Advisor
suppose our set is X = {x1,x2,...,xn}

consider the set of all products x1xi.

these have to be all distinct, since if:

x1xi = x1xj, by (left) cancellation we have: xi = xj

so for SOME xi, we must have x1xi = x1.

in a similar fashion (from right cancellation), we know that all products xjx1 are distinct.

so for any element in X, say xk, we know that xk = xjx1, for some xj in X.

so, if xi is a particular element of X with x1xi = x1, then for any xk in X:

xkxi = xjx1xi = xjx1 = xk

this means that xi is a right-identity for X.

now, for any xk in X, consider all products xkxj in X. again, these products are all distinct, so one of these products is xi, our right-identity.

that means every xk in X has some xj with xkxj = xi.

thus X has a right-identity and right-inverses.

now, note that xixk = (xixi)xk (since xi is a right-identity)

= xi(xixk) by associativity. so by left-cancellation:

xk = xixk, that is, xi is a left-identity as well. this means that xi is the (unique) identity for X.

now if xj is a (who knows, we might have more than one) right-inverse for xk, then:

xj = xjxi= xj(xkxj) = (xjxk)xj. but since we now know xi is also a left-identity,

xixj = (xjxk)xj, and by right-cancellation xi = xjxk, so xj is a left-inverse for xk as well.

since 2-sided inverses are necessarily unique, we have a group.

Thank you very much! Your explanation is splendid! :)