How to construct a semi-group over a non-numerical set? A semigroup is a set, e.g. $X$, with an associative binary operation, e.g. $\star$. That is for all $x,y,z \in X$, $(x \star y) \star z = x \star (y \star z)$.
I have seen some abstract algebraic books, but there are no examples for a non-numerical set. In other words, let $X = \{ a,b,c,d,e,f \}$. If we represent $\star$ by a table as follow
  *   a   b   c   d   e   f  
 ----------------------------
  a | . | . | . | . | . | . |
  b | . | . | . | . | . | . |
  c | . | . | . | . | . | . |
  d | . | . | . | . | . | . |
  e | . | . | . | . | . | . |
  f | . | . | . | . | . | . |
 ----------------------------

then what should I put in the table above instead of dots? In other words, what this matrix like representation should look like?
I am new to abstract algebra with a CS background. 
 A: Here's one way:
  *   a   b   c   d   e   f  
 ----------------------------
  a | a | a | a | a | a | a |
  b | a | a | a | a | a | a |
  c | a | a | a | a | a | a |
  d | a | a | a | a | a | a |
  e | a | a | a | a | a | a |
  f | a | a | a | a | a | a |
 ----------------------------

and as it says in one of the comments there are 15972 other semigroups with 6 elements (up to isomorphism and anti-isomorphism. 
You said: 

But I am generally interested to know the properties, if there are any in general, of this tabular representation so that (X,⋆) is a semigroup! 

One answer is the obvious one: the multiplication defined by the table has to be associative. 
Another answer might be: there is a result in this paper stating that almost all semigroups of order $n$ are 3-nilpotent, where almost all means that the proportion of semigroups with this property tends to $1$ as the order tends to infinity. 
A semigroup $S$ is 3-nilpotent if there exists an element $0\in S$ such that $xyz=0$ for all $x,y,z\in S$. 
One reason there are so many semigroups with this property is that their multiplication tables are very easy to define:


*

*Choose a zero element, let's say $e$ in the example with 6 elements;

*Choose a set of elements including $e$, say, $\{a,b\}$ in our example, and make every product of these elements equal to the zero element $e$;

*Define the product of any other two elements to be equal to any of the elements from step 2. 


The object with multiplication defined by this construction is always a semigroup, since the product of any 3 elements, written $x(yz)$ or $(xy)z$ is the zero element. 
For example,
  *   a   b   c   d   e   f  
 ----------------------------
  a | e | e | a | a | e | a |
  b | e | e | a | a | e | a |
  c | a | a | a | a | e | b |
  d | a | a | a | a | e | a |
  e | e | e | e | e | e | e |
  f | a | b | b | a | e | a |
 ----------------------------

defines an associative multiplication on the set $\{a,b,c,d,e,f\}$. 
The number of 3 nilpotent converges to the number of semigroups (up to isomorphism and anti-isomorphism) quite quickly:
$$
\begin{array}{l|l|l|l|l|l|l|l|l}
n&1&2&3&4&5&6&7&8\\\hline
\text{number of semigroups}&1& 4& 18& 126& 1160& 15973& 836021& 1843120128\\
 \text{number of 3-nilpotent semigroups}& 0& 0& 1& 8& 84& 2660& 609797& 1831687022\\
\end{array}
$$
So by $n=8$, 99% of the semigroups are 3-nilpotent. 
