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Prove that this finite set is a group

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    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 [itex]\in[/itex] G, the set <a> = {[itex]a^k[/itex] : k [itex]\in[/itex] N} is a finite closed subset of G. So, [itex]\exists[/itex]([itex]k_1[/itex], [itex]k_2[/itex])[itex]\in[/itex]N, such that:
    [itex]a^{k_1}=a^{k_2}[/itex] using the cancellation property I found that [itex]a=a.a^{k_2-k_1}[/itex].
    [itex]a^{k_2-k_1}[/itex] 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,
  2. jcsd
  3. Oct 28, 2011 #2


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    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.
  4. Oct 28, 2011 #3
    Thank you very much! Your explanation is splendid! :)
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