Mathematical Systems Question Help

Question

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?.

Answer 1

• 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 $c \ne a$.

Answer 2

As we can see through the table below:

• According to the table if we want to have an identity element, we should see one of the following cases:

          a    b    c
       ---------------
     a .  a    b    c
     b .  b
     c .  c
  

  Or:

          a    b    c
       ---------------
     a .       a     
     b .  a    b    c
     c .       c
  

  Or:

          a    b    c
       ---------------
     a .            a
     b .            b
     c .  a    b    c
  

so we see the second form above is coincide the table. It is not difficult that $b$ plays a role of an identity here. Just check the definition you had written for elements $a, b$ and $c$.

• If $id=b$ as it happened above, so search for an element $x$ among $a,b,c$ such that $$a*x=b,~~x*a=b$$ It is really $x=a$ and $x=c$. This shows that the structure is not a Group since the inverse is unique in any Group and for any element in it. Now do the same way for $b$ and $c$ to find additional information.

Answer 3

To demonstrate that you have an identity here, you need to show that for all a, there exists an e such that (aXe)=a=(eXa). So, you need to list a set of equations which show this true. Some of the equations you need are (bXa)=a, (bXb)=b, and (bXc)=c. Do you see the other ones you need to list also?

To spot an identity element, if it exists, on a table look at the leftmost column. Then look at the other columns and see if any of them matches the leftmost column (outside of the cross). Call this column C. The element at the top of the column in the cross can be the identity element, so long as it exists. An identity element will exist if there exists a similar row.

To show closure you just need to show that for all x, for all y, (xXy) is a member of the same set that x and y belong to. Since x and y in your example come from the set {a, b, c}, this means that you just need to show that for all x, for all y, (xXy) belongs to {a, b, c}.

To show commutativity you need to show that for all x, for all y, (xXy)=(yXx). So since x and y belong to {a, b, c}, that means you need to show that (aXb)=(bXa), (aXc)=(cXa), and (bXc)=(cXb). You can usually fairly easily tell if commutativity holds or not if there exists a symmetry over the diagonal.

To show that an inverse exists, you need to show that for all x, there exist a y such that (xXy)=e=(yXx), where e is the identity element of the system. There exists a geometrical way to spot if an inverse element exists, given an identity element. Here that means you need to list equations for a, for b, and for c, such as (aXc)=b=(cXa). If the system commutes, this cuts down the number of equations you need to show.

A group consists of a mathematical system (G, X) where G is the carrier set, and X a binary function such that

1. G satisfies closure under function X.
2. The system has an identity element.
3. For all elements of the system, there exists an inverse.
4. Associativity holds for the system, or that for all x, for all y, for all z (xX(yXz))=((xXy)Xz).

So you need to show that 1.-4. hold for your system if it's a group.

This requires (3x(3x3))=27 cases to check for your system for associativity. You can use something similar to truth tables to simplify the work here. Suppose we have a set {a, b}, and operation Y such that (aYb)=(bYa)=(aYa)=a and (bYb)=b. But, let's put operations at the end of expressions instead of in the middle. So, we have abY=baY=aaY=a and bbY=b. Associativity now means that for all x, y, z xyzYY=xyYzY. Then the following table shows that associativity holds:

 x  y  z  x yzY Y  xyY z Y
 a  a  a      a a    a   a 
 a  a  b      a a    a   a
 a  b  a      a a    a   a
 a  b  b      b a    a   a
 b  a  a      a a    a   a
 b  a  b      a a    a   a
 b  b  a      a a    b   a
 b  b  b      b b    b   b

since the final column for xyzYY and xyYzY match.