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.

Question

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.

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

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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:

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

$a*d = a*c$

$a*d = a = d*a$

and we end up with:

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

$b*d = b*c$

$b*d = b = d*b$

and we end up with:

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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:

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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.

Answer 1

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 $b$.