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Please help if there are any errors in this proof.

If a binary operation of multiplication on a nonvoid set $G$ is associative, and if all equations $xa=b$ and $ay=b$ have solutions $x$ and $y$ in $G$, then $G$ is a group.

Solution: First pick any element a' element of G. Then there exists an element u of G such that a'u= a' which is a solution to the equation. We must show that u is a left identity for all elements of G.

First we show left cancellation is possible. If there are two solutions for ax=b and ax'=b then ax=ax'and we must conclude that x=x'. Since if not x=x' then G cannot be a group. (Please help with this step in the proof, I am not convinced by its reasoning).

Now for all b of G, we have a'(ub)=a'b, using cancellation we have ub=b for all b of G, so u is a left identity. Use the equation xa= b for xa= u and this implies all a of G have a left inverse. These two properties makes G a group

Alex77
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  • Going from $ax=ax'$ to $x=x'$ presupposes that left cancellation holds. Further, building a contradiction with $G$ being a group is not a valid proof technique, since you are trying to prove that $G$ is a group. – vadim123 Nov 21 '15 at 05:49

1 Answers1

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Let $a\in G$ be arbitrary and $u$ with $au=a$, as above. Let $b\in G$. Then there is some $c\in G$ such that $ca=b$. We multiply $au=a$ on the left by $c$ to get $$bu=(ca)u=c(au)=ca=b$$ Hence $u$ is a right identity on all of $G$. By a similar, but mirror-reversed, process we can get $v$, a left identity on all of $G$. But now we apply each of these two identity properties to conclude that $$v=vu=u$$ Hence there is a two-sided identity. Now, for $a\in G$ the two properties guarantee us a left and a right inverse, i.e. $b,c\in G$ where $ba=u=ac$. But now we do the same trick again $$b=bu=b(ac)=(ba)c=uc=c$$ so in fact we have two-sided inverses.

vadim123
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