Anticommutative Operation with Left Cancellative Element is Right Operation -- why?

Question

Seth Warner's "Modern Algebra" (1965) once again: Exercise 7.17.

We are given a semigroup $(S, \circ)$ such that:

• $(1): \quad \forall x, y \in S: x \circ y = y \circ x \implies x = y$

• $(2): \quad \exists z \in S: \forall x, y \in S: z \circ x = z \circ y \implies x = y$

That is:

• $\circ$ is anticommutative

• $S$ has a left cancellative element.

The object of the exercise is:

Prove that $\circ$ is what Warner calls the "right operation", that is:

$\forall x, y \in S: x \circ y = y$

(which seems to me the same thing as saying that $x \circ y$ is the second projection on $(x, y)$, although this is probably going to be excoriated as appalling abuse of notation.)

I have tried various approaches, like:

• assuming $x \circ z = z \circ x$ and seeing where that takes me (to $z = x$ and where from there?)
• starting from the end: $x \circ y = y$, and working backwards via $z \circ (x \circ y) = z \circ y$ which is derived from $(z \circ x) \circ y = z \circ y$, but trying to get there from the axioms escapes me.

Again, a hint will probably be all I need.

Answer 1

First, note that anti-commutativity implies each element is idempotent, as $$(x \circ x) \circ x = x \circ (x \circ x) \implies x = x \circ x.$$ Next, we show $z \circ y = y$. We have, $$z \circ y = y \impliedby z \circ y \circ y = y \circ z \circ y \impliedby z \circ z \circ y \circ y = z\circ y \circ z \circ y \impliedby z \circ y = z \circ y,$$ the rightmost implication being true by idempotency. We also have $$x \circ z = z \impliedby z \circ x \circ z = x \circ z \circ z \impliedby x \circ z = x \circ z.$$ We can therefore conclude from this that any $x$ is left-cancellative: $$x \circ v = x \circ w \implies x \circ z \circ v = x \circ z \circ w \implies z \circ v = z \circ w \implies v = w.$$ Thus, the proof that $z \circ y = y$ works for any $x$, i.e. $x \circ y = y$ for all $x, y$.

Answer 2

Since $a$ and $a\circ a$ commute, we can conclude $a=a\circ a$, i.e. all $a\in S$ are idempotent.

(i) $z$ idempotent implies $z\circ z\circ x=z\circ x$, and left-cancelling $z$ gives $z\circ x=x$ for all $x\in S$.

(ii) If ever $x=x\circ z$ then we could replace the $x$ on the left side with $z\circ x$ to get $z\circ x=x\circ z$ which would imply that $x=z$. Therefore, $x\ne x\circ z$ for all $x\in S\setminus\{z\}$, i.e. the only fixed point of right-multiplication by $z$ is $z$ itself. But $x\circ z=x\circ z\circ z$ implies $x\circ z$ is a fixed point, so $x\circ z=z$ for all $x\in S$.

Finally, $x\circ y=x\circ (z\circ y)=(x\circ z)\circ y=z\circ y=y$. That is, $x\circ y=y$.

Answer 3

Here is my own attempt:

By associativity:

• $\forall a \in S: (a \circ a) \circ a = a \circ (a \circ a)$

By anticommutativity:

• $\forall a \in S: a \circ a = a$ that is idempotence.

Then by idempotence:

• $\forall a, b \in S: (a \circ a) \circ b \circ a = a \circ b \circ (a \circ a)$

By associativity:

• $a \circ (a \circ b \circ a) = (a \circ b \circ a) \circ a$

By anticommutativity:

• $(1): \quad a \circ b \circ a = a$

Now we have from $(1)$:

• $a \circ (c \circ b) \circ a \circ b = a \circ b$

and also from $(1)$:

• $a \circ c \circ (b \circ a \circ b) = a \circ c \circ b$

So:

• $\forall a, b, c \in S: a \circ c \circ b = a \circ b$

In particular, setting $a = z, c = x, b = y$:

• $\forall x, y, z \in S: z \circ x \circ y = z \circ y$

As we have asserted that $z$ is left cancellative:

• $\forall x, y \in S: x \circ y = y$

Ta-da.

The nudge that sent me down this route was the obvious fact of idempotence of an anticommutative associative operation (the first 2 lines).

Then I remembered those two crucial identities (included and proved in the above) that Warner had us prove back in Exercise $2.17$ of "Modern Algebra" where (now I look more closely) he actually refers us back to for that definition of anticommutativity:

• $\quad a \circ b \circ a = a$

and in particular as a result of this:

• $\forall x, y, z \in S: x \circ y \circ z = x \circ z$