Prove that if the cancellation law holds from both sides in a set, it forms a group.

Let G be a finite nonempty set with an operation * such that:

1. G is closed under *.

2. * is associative.

3. Given a,b,c in G with a*b=a*c, then b=c.

4. Given a,b,c in G with b*a=c*a, then b=c.

Prove that G must be a group under *.

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It's obvious that identity element satisfies the conditions 3 and 4, but I don't know whether that proves that the identity element is contained in G or not? moreover, How can I show that the inverse of any element in G is contained in G?

Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro

Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro

Quote:

Originally Posted by

**FernandoRevilla** Let

. Suppose

where

are distinct. Use the left cancellation law to prove that

Would you explain how you obtained this set?

Quote:

This implies that the equation

is solvable . Similarly we can prove that the equation

is solvable . Could you continue ?

This is how I understood your post: since for any a,b in G is solvable, then by definition there exists an x that satisfies that relation, so if let a=b, that proves the existence of , the same logic can be deployed to prove the existence of , now it'll be easy to prove that these two are equal. after proving that the left and right identity elements are the same and we denote the identity element by e, now if we let , since it's solvable, it proves the existence of , the same logic can be applied to the equation and that proves the existence of , now again it's easy to verify that these two will be the same element.

Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro

Quote:

Originally Posted by

**FernandoRevilla** Let

. Suppose

where

are distinct. Use the left cancellation law to prove that

.

Quote:

Originally Posted by

**Nikita2011** Would you explain how you obtained this set?

Well, the elements of this set are explicitly enumerated. We choose some , multiply it on the right by every element of and collect the results in a new set . The important claim about is that because is closed under *. Further, the function ; is an injection (why?), so and is a bijection. Since is in particular a surjection, for every there exists an such that .

Quote:

Originally Posted by

**Nikita2011** This is how I understood your post: since for any a,b in G

is solvable, then by definition there exists an x that satisfies that relation, so if let a=b, that proves the existence of

Yes, but not immediately. You have a solution to the equation for some particular . However, you need to show for all .