Prove that this finite set is a group

In summary, a nonempty finite set with an associative binary operation satisfying left and right cancellation is proven to be a group by showing that it has a unique identity and every element has a left and right inverse.
  • #1
nata
2
0

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
 
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  • #2
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
Thank you very much! Your explanation is splendid! :)
 

1. What is a finite set?

A finite set is a set that contains a limited or countable number of elements. This means that the number of elements in the set is finite and can be counted or listed.

2. What is a group?

A group is a mathematical structure that consists of a set of elements and a binary operation that combines any two elements to form a third element. The operation must also satisfy certain properties, such as associativity, identity, and inverse elements.

3. How do you determine if a finite set is a group?

To determine if a finite set is a group, you need to check if the set satisfies the four group properties: closure, associativity, identity, and inverse elements. This means that for any two elements in the set, when the binary operation is applied, the result must also be an element in the set. The operation must also be associative, there must be an identity element, and each element must have an inverse element.

4. Can a finite set be a group if it does not have an identity element?

No, a finite set cannot be a group if it does not have an identity element. The identity element is a crucial component of a group, as it ensures that every element in the set has an inverse element and that the group is closed under the operation.

5. What are some examples of finite sets that are also groups?

Examples of finite sets that are also groups include the set of integers under addition, the set of non-zero real numbers under multiplication, and the set of symmetries of a square under composition. Any finite set that satisfies the four group properties can be considered a group.

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