# Fill in a partly filled in table such that it makes the magma $(M,*)$ associative, commutative, has an identity element and has no zero-elements.

Below is a partly filled in table for a binary operation ($$*$$) on the set $$M=\{a,b,c,d\}$$. I am trying to fill in the rest such that the magma $$(M,*)$$ becomes associative, commutative, has an identity element and has no zero-elements.

Using the fact that $$(M,*)$$ is supposed to be commutative we can fill in a few cells and get:

Now we want the magma to be associative too, so I can get the following information:

$$(c*a)*b = c*(a*b)$$

$$a*b = c*b$$

$$b*c = b = c*b$$

and we end up with:

$$(a*c)*d = a*(c*d)$$

$$a*d = a*c$$

$$a*d = a = d*a$$

and we end up with:

$$(b*c)*d = b*(c*d)$$

$$b*d = b*c$$

$$b*d = b = d*b$$

and we end up with:

Now we see that $$d$$ must be the identity element due to the d-row being equal to the heading row and the d-column is equal to the heading column.

we end up with:

Now this is where I get stuck, I can't figure out what the two last cells need to be. I have gained some information though, the magma $$(M,*)$$ does not form a group. This due to the fact that we for example have duplicate entries on the a-row [_ b a a] which would not be allowed if the magma was a group. We also notice that not every element seems to have an inverse, which again is not allowed if the magma was to be a group.

• I guess it's because there are multiple options. Oct 15, 2020 at 15:41
• Indeed, probably yes. We have $a*a=a*c*c=a*c=a$. If there's no zero element, then $b*b\ne b$. Oct 15, 2020 at 15:47
• The most immediate inference is that $b*b\ne b$. Oct 15, 2020 at 15:49
• @NoName123: It’s an absorbing element: an element $z$ with the property that $z*a=z$ for all $a$. $0$ is such an element for ordinary multiplication. Oct 15, 2020 at 16:50
• @NoName123: Sorry: I misread your comment, and you should ignore mine, which I’ll delete. Oct 15, 2020 at 17:04

Since $$ac = a$$ and $$a =c^2$$, you have $$a^2 =aa=ac^2 = ac = a$$. I claim that $$b^2 = a$$. Indeed, if $$b^2 = 1$$, then since $$ab = b$$, $$1 = b^2 = (ab)^2 = a^2b^2 = a^2$$, a contradiction, since $$a^2 = a$$. If $$b^2 = c$$, then since $$ab =b$$, one has $$abb = bb$$, that is, $$ac = c$$, a contradiction, since $$ac = a$$. Finally, if $$b^2 = b$$, then $$b$$ is a zero of the monoid, which is excluded. This proves the claim.
Altogether the identity of the monoid is $$1$$, there is a non-regular element $$c$$ and the minimal ideal is the group of order $$2$$ consisting of the idempotent $$a$$ and the element $$b$$.
• What do you mean with "the idempotent $b$"? Shouldn't an idempotent be a member $x$ such that $x^2=x$? Oct 17, 2020 at 18:09
• @amsra Ooops, you're right: $a$ is the idempotent and $b$ is the other element of the group. Fixed. Oct 17, 2020 at 19:56