Prove that this finite set is a group

[nata]

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)\inN, 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

 

[Deveno]

suppose our set is X = {x₁,x₂,...,x_n}

consider the set of all products x₁x_i.

these have to be all distinct, since if:

x₁x_i = x₁x_j, by (left) cancellation we have: x_i = x_j

so for SOME x_i, we must have x₁x_i = x₁.

in a similar fashion (from right cancellation), we know that all products x_jx₁ are distinct.

so for any element in X, say x_k, we know that x_k = x_jx₁, for some x_j in X.

so, if x_i is a particular element of X with x₁x_i = x₁, then for any x_k in X:

x_kx_i = x_jx₁x_i = x_jx₁ = x_k

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

now, for any x_k in X, consider all products x_kx_j in X. again, these products are all distinct, so one of these products is x_i, our right-identity.

that means every x_k in X has some x_j with x_kx_j = x_i.

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

now, note that x_ix_k = (x_ix_i)x_k (since x_i is a right-identity)

= x_i(x_ix_k) by associativity. so by left-cancellation:

x_k = x_ix_k, that is, x_i is a left-identity as well. this means that x_i is the (unique) identity for X.

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

x_j = x_jx_i= x_j(x_kx_j) = (x_jx_k)x_j. but since we now know x_i is also a left-identity,

x_ix_j = (x_jx_k)x_j, and by right-cancellation x_i = x_jx_k, so x_j is a left-inverse for x_k as well.

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

 

[nata]

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