Explaining Why a Set with Operation * Does Not Define a Group

[coderot]

Hi,

I'm having trouble understanding why the follow composition table for the set \left\{ a, b, c, d \right\} with operation * doesn't define a group.
<br /> \begin{array}{c|cccc}<br /> * &amp; a &amp; b &amp; c &amp; d \\ \hline<br /> a &amp; c &amp; d &amp; a &amp; b \\ <br /> b &amp; d &amp; c &amp; b &amp; a \\ <br /> c &amp; a &amp; b &amp; c &amp; d \\ <br /> d &amp; b &amp; a &amp; d &amp; c \\ <br /> \end{array}<br />
Firstly I know that the operation is closed since every element in the set is in the table. The operation is commutative because it's symmetrical about the leading diagonal and the identity element is c (the third row).

However according to the example the operation isn't associative. This is what I'm having trouble with. According to the book (New Comprehensive Mathematics for 'O' Level) the example says that b * (d * a) = b * b = b and this is what I don't understand. Why is the result of this operation not c?. From the table is says that b * b = c. Any help please thanks.

 

[pbuk]

It looks like there is a misprint. As you say, from the table b * b = c and as (b * d) * a = a * a = c this not an example of non-associativity.

 

[coderot]

That is what I thought. However there is another example of this in the accompanying exercises. Here is the table. I'm asked to state why this isn't a group.
<br /> \begin{array}{c|ccc}<br /> * &amp; 0 &amp; 2 &amp; 4 \\ \hline<br /> 0 &amp; 0 &amp; 2 &amp; 4 \\ <br /> 2 &amp; 2 &amp; 0 &amp; 2 \\ <br /> 4 &amp; 4 &amp; 2 &amp; 0 \\<br /> \end{array}<br />
Again the result states that the set \left\{0, 2, 4\right\} isn't a group for this operation because is isn't associative. Which makes me believe that I'm not understanding something about this property of groups.

 

[mathman]

That is what I thought. However there is another example of this in the accompanying exercises. Here is the table. I'm asked to state why this isn't a group.
<br /> \begin{array}{c|ccc}<br /> * &amp; 0 &amp; 2 &amp; 4 \\ \hline<br /> 0 &amp; 0 &amp; 2 &amp; 4 \\ <br /> 2 &amp; 2 &amp; 0 &amp; 2 \\ <br /> 4 &amp; 4 &amp; 2 &amp; 0 \\<br /> \end{array}<br />
Again the result states that the set \left\{0, 2, 4\right\} isn't a group for this operation because is isn't associative. Which makes me believe that I'm not understanding something about this property of groups.

(2x4)x4 = 2x4 = 2
2x(4x4) = 2x0 = 0

 

[Stephen Tashi]

Another example: Try (2*2)*4 vs 2*(2*4).

In an actual group, the equation g*X = h has the unique solution X = g^{-1} *h
In this set, the equation 2*X = 4 has no solution.
In a real group you can multiply an equation on both sides by the same group element without changing the solution set.

So 2*X = 4 would have the solution X = 2^{-1}*4 = 2*4

Substituting this back into the equation 2*X = 4 we have
2*(2*4) = 4 which suggested to me that there is some problem with evaluating the product 2*2*4.

 

[coderot]

Thanks for your responses guys. You've cleared this one up for me. :)