Solutions to equations in a group: Interpretation inquiry

Question

I'm working on the following question:

Show that $G$ is a group if and only if, for every $a, b \in G$, the equations $xa = b$ and $ay = b$ have solutions $x, y \in G$.

I'm having trouble getting started because I'm not understanding what it means for "the equations $xa = b$ and $ay = b$ have solutions $x, y \in G$". Do they mean, there's only 1 left multiplier that takes $a$ to $b$ and one right multiplier that takes $a$ to $b$?

Answer 1

I think our colleagues stressed out and ACTOH have made clear what we're looking for here.

I would phrase it like this:

Let $G \ne \emptyset$ be a set with an associative binary operation; then $G$ is a group if and only if for any $a, b \in G$ there exist $x, y \in G$ such that $xa = b$ and $ay = b$.

The details of the proof of this little proposition are in and of themselves worth seeing; basic and classic stuff.

The "only if" direction is pretty obvious, so I'll focus on the "if" direction here.

So suppose

$xa = b \tag 1$always has a solution. Then taking

$b = a \tag 2$

we have some $e_l$ such that

$e_l a = a; \tag 3$

I claim that in fact

$\forall b \in G, \; e_l b = b; \tag 4$

for since

$\exists y \in G, \; ay = b, \tag 5$

we have

$e_l b = e_l (ay) = (e_la)y = ay = b; \tag 6$

likewise, by reversing more or less the roles played by the equations $xa = b$ and $ay = b$, we find the existence of an $e_r \in G$ such that

$\forall b \in G, \; be_r = b; \tag 7$

then

$e_l = e_l e_r = e_r; \tag 8$

taking

$e = e_l = e_r, \tag 9$

we have our multiplicative identity for $G$. We proceed: taking $b = e$ in the equation $xa = b$; then there is an $a' \in G$ with

$a'a = e; \tag{10}$

likewise, from $ay = b$, again with $b = e$, we have $a'' \in G$ such that

$aa'' = e; \tag{11}$

thus,

$a'' = ea'' = (a'a)a'' = a'(aa'') = a'e = a'; \tag{12}$

we may now, in the light of (10)-(12), take

$a^{-1} = a' = a'', \tag{13}$

and we have found the inverse for $a \in G$.

Since we now have the associative binary operation on $G$ is possessed of an identity $e \in G$, and that every $a \in G$ is possessed of an inverse element $a^{-1}$, we conclude that $G$ is indeed a group.

Answer 2

"the equations $xa = b$ and $ay = b$ have solutions $x,y \in G$", means precisely this :

For all $a \in G$ and $b \in G$, there exist $x \in G$ and $y \in G$ which satisfy the equations $xa = b$ and $ay = b$ respectively.

Since it only says that there is at least one solution to $xa = b$, the hypothesis does not rule out $x'a = b$ or $ay' = b$ for some other $x',y' \in G$. It is true that this will hold if $G$ is a group (by multiplying by the left/right inverse of $a$), but you cannot assume this from hypothesis.

Note : While it is not mentioned, $G$ must have a binary product on it that is associative for this to work out. (and must be non-empty, as pointed out in the comment below)

Answer 3

What it means is as follows: $$\forall a,b \in G, \exists x \in G: xa=b$$ $$\forall a,b \in G, \exists y \in G: ay=b$$

Assuming that $(G,\star)$ is a semi-group, i.e. $G \not= \emptyset$ is closed and associative under $\star$, you must prove the existence of the identity element and the existence of the inverse element for each element in $G$.

Hint:

First show that for each element $a \in G$, one can find $e(a)_r$ and $e(a)_l$ such that $e(a)_l\star a = a \star e(a)_r = a$. Then show that $e(a)_r=e(a)_l:=e(a)$ is independent of $a$, i.e. for any element $a$, one can take $e(a)$ to be a fixed unique element $e \in G$.

Now using the same kind of argument, show that for each element $a \in G$, you can find an inverse for $a$. Again, this means that you should prove the existence and equality of the following elements: $a^{-1}_l=a^{-1}_r := a^{-1}$