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Prove that this finite set is a group

  • Thread starter nata
  • Start date
  • #1
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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 [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].
So,
[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,
NAta
 

Answers and Replies

  • #2
Deveno
Science Advisor
906
6
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.
 
  • #3
2
0
Thank you very much! Your explanation is splendid! :)
 

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