Group problem with a diagram

Question

Let the set $M=\{a,b,c,d\}$ and a binary operation $\star$ described in the diagram :

$$\begin{array}{c|cccc} \star & a & b & c & d \\ \hline a & a & b & a & b \\ b & b & a & b & a \\ c & a & b & c & d \\ d & b & a & d & c \\ \end{array}$$

Questions given: Is $(M,\star)$ a group? Is $\star$ commutative? Is $c$ the neutral element of $\star$? Is $(\{a,b\},\star)$ a group? Is $(\{a,c\},\star)$ a group?

I obtained that $c$ is the neutral element of the group from the diagram and also that the binary operation is commutative but I am not sure about the other statements.

Answer 1

In a group there is a single element $g$ such that $g^2=g$ (why?). However, your structure has two such elements, $a$ and $c$, and so is not a group

For the other two questions: the above also means that $\{a, c\}$ is not a group. On the other hand, $\{a,b\}$ is a group, and proving this is an instructive exercise :-)

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We can use the table to find the neutral element easily: A neutral element is an element such that the corresponding row and column in the table are precisely the outer row and column, respectively (so both are $a,b,c,d$). There can be at most one neutral element. Therefore, $c$ is the neutral element.

We can also see commutativity via the table: this corresponds to the table being symmetric, with the line of symmetry going from top left to bottom right. This is the case here, so the structure is also commutative.

Answer 2

It is not a group, for it does not satisfy the Latin Square property: the first row has two $a$s.

Answer 3

It is not a group because at it hasn't the simplification property. Indeed we have :

$$a*a = a*c$$

and "left-simplifying" by $a$, we obtain $a=c$ which is not true.

Remark: in fact, this property is equivalent to the "latin table" property but more usable in the case the table is not explicitly given.

Answer 4

We can also explicitly describe this structure. It turns out it is a monoid. It is the product of the monoid $(\{0,1\}, \max)$ and the group $(\mathbb{Z}_2, +)$.

We have $$ a = (1, 0) \\ b = (1, 1) \\ c = (0, 0) \\ d = (0, 1) $$

It is easy to see that $a$ does not have an inverse, that $c$ is the neutral element, that $\{a, b\}$ is a group (with a different neutral element, $a$) and that $\{a, c\}$ is not a group ($a$ has no inverse).

Your structure is furthermore a semiring $(\{0,1\}, \max, \cdot) \times (\mathbb{Z}_2, +, \cdot)$.