# Proving a binary operation on $\mathbb{Z}$ gives a semigroup

Let $x\circ y= x +y-xy, \quad (x,y) \in \mathbb{Z}$

where $\circ$ is a binary operation on $\mathbb{Z}$, prove that this is a semigroup.

My Work

To prove we have to check two things:

• $\mathbb{Z}$ is closed under $\circ$

• $\circ$ is associative on $\mathbb{Z}$

For the first one, since we are only adding, subtractive and multiplying integers, the results will always be integers. Hence $\mathbb{Z}$ is closed under $\circ$.

For the second, I just have to show that $(x\circ y) \circ z =x\circ(y\circ z)$.

RHS: $(x\circ y) \circ z =(x+y-xy)\circ z= x+y-xy+z-zx-zy+xyz$

LHS: $x\circ(y\circ z)=x\circ (y+z-yz)=x +y-xy+z-zx-zy+xyz$

Hence RHS$=$LHS.

Therefore $(\mathbb{Z},\circ)$ is a semigroup.

Is this correct? And is there a better way to show that it's closed?

-
Looks right to me. Usually these elementary algebra exercises are just plug-and-chug, nothing flashy about the proofs. – BaronVT Mar 13 '13 at 4:57

This structure as @Cameron confirmed, defines a infinite semigroup $S$. $S$ has $0$ and $1$ as two idempotent elements, since $0*0=0$ and $1*1=1$ and so it is not a band. It seems that $S$ has $0$ as its identity $e$ i.e. for all $a\in S$, $a*e=e*a=a$. $S$ looks like a commutative semigroup, since $$x*y=y*x, \forall x,y\in\mathbb Z$$

-
Nice...nothing left for me to add! +1 – amWhy Mar 13 '13 at 5:23
$x \circ 0 = x$ for all $x$, please see my answer. – Andreas Caranti Mar 13 '13 at 6:22
@AndreasCaranti: Yes! I lost it. Thanks. – Babak S. Mar 13 '13 at 6:57

Perhaps it is instructive to see how these exercises arise. This also gives a handy solution: once you have the general setup described below, there is really nothing to prove.

Consider the map $\varphi : \Bbb{Z} \to \Bbb{Z}$ given by $\varphi(x) = 1 - x$.

Note that $\varphi^{2} = \mathbf{1}_{\Bbb{Z}}$, so that $\varphi$ is bijective, and $$\varphi(x) \varphi(y) = (1 - x) (1 - y) = 1 - x - y + xy = 1 - (x + y - xy) = \varphi(x \circ y),$$ so that $$x \circ y = \varphi^{-1}(\varphi(x) \varphi(y)).$$
(Ok, I know $\varphi^{-1} = \varphi$ in this particular case.) So $(\Bbb{Z}, \circ)$ is obtained through transport of structure from $(\Bbb{Z}, \cdot)$, and $$\varphi: (\Bbb{Z}, \circ) \to (\Bbb{Z}, \cdot)$$ is an isomorphism. The structure of a monoid of $(\Bbb{Z}, \cdot)$ carries over to $(\Bbb{Z}, \circ)$ verbatim, you are really verifying the familiar axioms in $(\Bbb{Z}, \cdot)$, only indirectly via $\varphi$. In particular, the identity of $(\Bbb{Z}, \circ)$ is $\varphi^{-1}(1) = 0$, where $1$ is the identity of $(\Bbb{Z}, \cdot)$.

By the way, this circle operation (or a slight variation thereof) plays an important role in the theory of the Jacobson radical.

-
+1 Thanks for remarking the defect in the answer. – Babak S. Mar 13 '13 at 6:59
@BabakS., you're very welcome. It's just that once you get the isomorphism, things are so much easier to see. – Andreas Caranti Mar 13 '13 at 7:33

Your method of showing closure is pretty much as good as it gets, and your associativity proof is just fine.

-
Okay, Thank you! – MITjanitor Mar 13 '13 at 4:59