Mathematical Systems Question Help Alright, this is another question for my math for teachers course. This question is not actually in the homework, but there are problems similar to it. I'd really like to learn how to do problems like this one, so it would be much appreciated if I was given a solid explanation/demonstration. I don't really have a clue where to start, so I've defined the terms as best as I can under each question... The question should be answered using the data from the table.
A mathematical system is defined by the following table:

Is there an identity element in this system? If so, what is it?
Identity is: aXe=a and eXa=a
Is closure satisfied by this system? Explain.
Closure is: aXb
Is this system commutative? Explain.
Commutative is: aXb=bXa
Does every element have an inverse? Explain.
Inverse is: aXx=e and xXa=e
Is this system a group? Explain.
Not sure what they mean by this... 
Are they referring to another property?. 


*

*Scanning along the rows, we see that the second row is in alphabetical order. Similarly, scanning along the columns, we see that the second column is in alphabetical order. Thus, $b$ must be the identity element.

*Since all elements in the grid are either $a,b,$ or $c$, we know that the system is closed under multiplication.

*Since the grid is symmetrical with respect to the main diagonal (draw a line going from the top left to the bottom right), the system is commutative.

*Since the identity element $b$ appears at least once in every row and column, every element has an inverse (note that inverses aren't necessarily unique, suggesting that it isn't a group).

*A group is a set $G$ with a multiplication operation $\times$ that satisfies 4 properties: identity, closure, inverses, and associativity. Since the first 3 conditions are satisfied, it suffices to consider the last condition:

Definition: An operation $\times$ is considered associative iff for all $x,y,z\in G$: $$ (x \times y) \times z = x \times (y \times z)$$
  
*Observe that this system is not associative (and thus not a group), since:
  $$
(a \times a) \times c=b \times c = c\\
a \times (a \times c)=a \times b = a
$$
  Yet $c \ne a$.
  

