Group problem with a diagram 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.
 A: 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 :-)

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.
A: It is not a group, for it does not satisfy the Latin Square property: the first row has two $a$s.
A: 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.
A: 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 $(\{0,1\}, \max, \cdot) \times (\mathbb{Z}_2, +, \cdot)$.
