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

• Nov 22nd 2011, 08:31 AM
Nikita2011
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?
• Nov 22nd 2011, 09:32 AM
FernandoRevilla
Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro
Quote:

Originally Posted by Nikita2011
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 *.

Let $\displaystyle a,b\in G$ . Suppose $\displaystyle G=\{a_1,\ldots,a_n\}$ where $\displaystyle a_1,\ldots,a_n$ are distinct. Use the left cancellation law to prove that $\displaystyle G=\{aa_1,\ldots,aa_n\}$ . This implies that the equation $\displaystyle ax=b$ is solvable . Similarly we can prove that the equation $\displaystyle ya=b$ is solvable . Could you continue ?
• Nov 26th 2011, 05:35 AM
Nikita2011
Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro
Quote:

Originally Posted by FernandoRevilla
Let $\displaystyle a,b\in G$ . Suppose $\displaystyle G=\{a_1,\ldots,a_n\}$ where $\displaystyle a_1,\ldots,a_n$ are distinct. Use the left cancellation law to prove that $\displaystyle G=\{aa_1,\ldots,aa_n\}$

Would you explain how you obtained this set?

Quote:

This implies that the equation $\displaystyle ax=b$ is solvable . Similarly we can prove that the equation $\displaystyle ya=b$ is solvable . Could you continue ?
This is how I understood your post: since for any a,b in G $\displaystyle ax=b$ is solvable, then by definition there exists an x that satisfies that relation, so if let a=b, that proves the existence of $\displaystyle e_R$, the same logic can be deployed to prove the existence of $\displaystyle e_L$, 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 $\displaystyle ax=e$, since it's solvable, it proves the existence of $\displaystyle a^{-1}_R$, the same logic can be applied to the equation $\displaystyle ya=e$ and that proves the existence of $\displaystyle a^{-1}_L$, now again it's easy to verify that these two will be the same element.
• Nov 26th 2011, 06:06 AM
emakarov
Re: Prove that if the cancellation law holds from both sides in a set, it forms a gro
Quote:

Originally Posted by FernandoRevilla
Let $\displaystyle a,b\in G$ . Suppose $\displaystyle G=\{a_1,\ldots,a_n\}$ where $\displaystyle a_1,\ldots,a_n$ are distinct. Use the left cancellation law to prove that $\displaystyle G=\{aa_1,\ldots,aa_n\}$ .

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 $\displaystyle a\in G$, multiply it on the right by every element of $\displaystyle G$ and collect the results in a new set $\displaystyle G'$. The important claim about $\displaystyle G'$ is that $\displaystyle G'\subseteq G$ because $\displaystyle G$ is closed under *. Further, the function $\displaystyle f:G\to G'$; $\displaystyle f(a_i)=aa_i$ is an injection (why?), so $\displaystyle G=G'$ and $\displaystyle f$ is a bijection. Since $\displaystyle f$ is in particular a surjection, for every $\displaystyle b\in G$ there exists an $\displaystyle x\in G$ such that $\displaystyle ax=b$.

Quote:

Originally Posted by Nikita2011
This is how I understood your post: since for any a,b in G $\displaystyle ax=b$ is solvable, then by definition there exists an x that satisfies that relation, so if let a=b, that proves the existence of $\displaystyle e_R$

Yes, but not immediately. You have a solution $\displaystyle e_R$ to the equation $\displaystyle ae_R=a$ for some particular $\displaystyle a$. However, you need to show $\displaystyle be_R=b$ for all $\displaystyle b\in G$.