Seth Warner's "Modern Algebra" (1965) presents in Chapter 7 a whole series of exercises which present the student with a semigroup with certain additional properties, and demonstrate that these properties lead to that semigroup being a group.
However, I have had no joy in getting "there" from "here", so to speak.
Exercise 7.16 goes like this:
"Let $(E, \cdot)$ be a commutative semigroup satisfying:
$1^\circ \quad$ For every $x \in E$ there exists $y \in E$ such that $y x = x$
$2^\circ \quad$ For all $x, y \in E$, if $y x = x$, then there exists $z \in E$ such that $z x = y$.
Show that:
- (a) If $y x = x = y' x$, then $y = y'$.
- (b) If $y x = x$, then $y y = y$.
- (c) If $y x = x$ and $z w = w$, then $y = z$.
- (d) $(E, \cdot)$ is a group."
Clearly we are being groomed to accept the fact that $y$ is the unique identity, and that the given $z$ in the conditions is the inverse of $x$.
I can't see which direction to take. I have munged together a whole series of group products from the set $\{x, y, y', z\}$ etc., but I can't see how to simplify what I get.
Any hints as to where to start? As from the previous question from this exercise set, it may be that a single judicious hint is enough to get the ball rolling, so to speak.
EDIT: From (a) I now have (b) and (c) (many thanks to Robert Shore and podiki), but I am still stumped on how to solve (a).
EDIT 2: (a) is also solved.
For (a):
Assume that $yx=y'x=x$. By 2, there exists $z$ with $zx=y$.
Now $yy' = zxy' = zy'x=zx =y$, and similarly $yy' = y'$, so $y=y'$.