Let $*$ denote the binary operation defined on the set $\Bbb Z$ of integers, where

$$x * y = 3xy - 5x - 5y + 10$$

for all integers $x$ and $y$. Prove that $\Bbb Z$, with the binary operation $*$, is a monoid. What is the identity of the monoid? Find all invertible elements of the monoid. Is the monoid a group?

I dont understand how to use $x*y*z$ to find if it is a monoid or not. after the first step: $$x * y * z = (3xy - 5x - 5y + 10)*z$$ I am confused as to what $z$ should equal for this equation.

  • $\begingroup$ You will get better help by writing out the image. $\endgroup$
    – user208649
    Apr 12 '16 at 17:52
  • $\begingroup$ ok, I wrote it out there $\endgroup$
    – Joe Daly
    Apr 12 '16 at 17:57
  • $\begingroup$ Ok, so just let z remain a variable, and apply the operation again. Also note that you should have written $(x*y)*z$ because you have not yet proven associativity. You want to verify that $(x*y)*z = x*(y*z)$. $\endgroup$
    – user208649
    Apr 12 '16 at 18:02
  • $\begingroup$ can you be more precise? What do you mean by "apply the operation again"? $\endgroup$
    – Joe Daly
    Apr 12 '16 at 18:07
  • $\begingroup$ You have an operation $*$, so just do it again with the RHS of your last equation. $\endgroup$
    – user208649
    Apr 12 '16 at 18:20

You must show that $*$ is associative, i.e., that


for all $x,y,z\in\Bbb Z$. A simple approach is to expand each side of $(1)$ separately and then verify that the expansions are equal. For example,

$$\begin{align*} (x*y)*z&=(3xy-5x-5y+10)*z\\ &=3(3xy-5x-5y+10)z-5(3xy-5x-5y+10)-5z+10\\ &=(9xy-15x-15y+30)z-15xy+25x+25y-50-5z+10\\ &=9xyz-15xz-15yz+30z-15xy+25x+25y-50-5z+10\\ &=9xyz-15xy-15yz-15xz+25x+25y+25z-40\;. \end{align*}$$

The first step here is just applying the definition of $*$ to $x*y$. We then treat the resulting integer, $3xy-5x-5y+10$, as the first operand of a similar expansion. If that’s a little hard to follow at first, temporarily let $u=3xy-5x-5y+10$; the second step of the computation is then simply rewriting $u*z$ as $3uz-5u-5z+10$. The rest of the computation is ordinary algebra.

Now expand $x*(y*z)$ into an ordinary algebraic expression in similar fashion; if you get the same expression, you’ve proved that $*$ is associative.

To find the identity, you must identify an integer $n$ such that $n*x=x*n=x$ for all $x\in\Bbb Z$. Actually, it’s pretty easy to prove that $*$ is commutative, and once you’ve done that, you need only check that $n*x=x$ for all $x\in\Bbb Z$. Thus, you want $n\in\Bbb Z$ such that


for all $x\in\Bbb Z$. Solving $(2)$ for $n$ is straightforward.

Once you know what the identity $n$ is, you want to find all invertible $x\in\Bbb Z$, i.e., all those for which there is a $y\in\Bbb Z$ such that $x*y=n$. For this I suggest using the definition of $*$ to write $x*y=n$ as an ordinary algebraic equation, solving that equation for $y$ in terms of $x$, and determing for what integer values of $x$ the solution $y$ is also an integer.

  • $\begingroup$ $(x*y)*z=9xyz-\dots$, and yes, it is associative. $\endgroup$
    – almagest
    Apr 13 '16 at 10:47
  • $\begingroup$ @almagest: It is indeed. (Sorry about the typo; I’m not sure how $9$ turned into $27$, but I’m fixing it now.) $\endgroup$ Apr 13 '16 at 20:01

The symbol $x$ when you write $x*y$ should be allowed to stand for any integer $x$, even one written in a complicated form. So to compute

$$ (3xy - 5x - 5y + 10) * z, $$

you take the formula for $x*y = 3xy - 5x - 5y + 10$ that you were given, and on the right side, substitute $3xy-5x-5y+10$ for 'x' and z for 'y'.


I will try to help you by clarifying what it means to apply an operation two times because it looks that you are stuck on that.

Let us use some example different from yours. Define operation $*$ on $\mathbb Z$ as $x*y=xy+x+y$.

Now to see what is $(x*y)*z$:


So we made substitution $w=xy+x+y$ and then applied definition of our operation on $w*z$ and then instead of $w$ wrote $xy+x+y$ to calculate what $(x*y)*z$ really is.

Now you can try in this way with your operation.


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