Prove that a nonempty set G, together with an associative binary operation * on G such that a∗x=b and y∗a=b have solutions in G, ∀a,b∈G , is a group.
I appreciate any help I can get.
Since a*x and y*a are both equal to b, they are equal to each other: a*x= y*a. Since this is a group, every member, including a, has an inverse. Multiplying both sides on the right by a gives $\displaystyle y= a*x*a^{-1}$. Choose any member of G to be x and calculate y.
Halls, he has to prove such a set is a group, he cannot assume the existence of an identity, or the existence of inverses.
Yeison, Since you know $G$ is non-empty, you know you can pick at least one element $a \in G$.
Thus we know the equation:
$a \ast x = a$ has a solution (the solution might be $a$ itself, or it might be some other element of $G$, if $G$ has any).
Right now, this $x$ seems the most likely candidate for the identity of $G$. But we don't know this, yet.
Now suppose that $b$ is any element of $G$. We know there is $y$ in $G$ such that $y \ast a = b$ (remember, $a$ here is some specific element of $G$, and so is $x$, and $b$ is arbitrary).
Now: $b \ast x = (y \ast a) \ast x = y \ast (a \ast x) = y \ast a = b$. This is true for ANY $b \in G$, so $x$ is a right-identity for $G$.
So we have some $x \in G$ (namely, the $x$ such that $a \ast x = a$) that functions as a right identity for ALL of $G$.
Now, we also have some (maybe more than one, but at least one) $y \in G$ such that $y \ast a = a$. Again, for any $b$, we have some $z \in G$ with:
$a \ast z = b$.
So (by the same trick as before): $y \ast b = y \ast (a \ast z) = (y \ast a) \ast z = a \ast z = b$.
So this (specific) $y$ functions as a left-identity for ALL of $G$.
So, now, the kicker:
$y = y \ast x = x$ (since $y$ is a left-identity, and $x$ is a right-identity).
So, our original $x$ and the $y$ are the SAME element of $G$, and form a two-sided identity. Let's call it $e$.
So we know $G$ has an identity. What about inverses?
Well, since the equation:
$a \ast x = b$ has a solution $x$ for EVERY $a,b \in G$, we can, for any $a \in G$ take $b = e$, so the equation:
$a \ast x = e$ has a solution. This $x$ (which is different than our previous $x$, sorry about that, but it's really easy to "run out of letters") is clearly a right-inverse for $a$.
Since we can do this for ANY $a \in G$, ALL elements of $G$ have at least one right-inverse (maybe more, we don't know, yet).
Similarly, since:
$y \ast a = e$ has a solution, every element $a$ of $G$ has a left-inverse. So let $y$ be a left-inverse for $a$, and let $x$ be a right-inverse for $a$.
Then:
$y = y \ast e = y \ast (a \ast x) = (y \ast a) \ast x = e \ast x = x$, so any left-inverse for $a$ is also a right-inverse for $a$.
Now, suppose that $x,x'$ are two two-sided inverses for $a$.
Then $x' = x \ast e = x' \ast (a \ast x) = (x' \ast a) \ast x = e \ast x = x$, so each $a \in G$ has a UNIQUE inverse. Thus, $(G,\ast)$ is a group.
See also math.stackexchange.