Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $$x*y=y*x, \forall x,y\in\mathbb Z$$

share|cite|improve this answer
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.

share|cite|improve this answer
+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.

share|cite|improve this answer
Okay, Thank you! – MITjanitor Mar 13 '13 at 4:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.