- #1
coderot
- 10
- 0
Hi,
I'm having trouble understanding why the follow composition table for the set [itex]\left\{ a, b, c, d \right\} [/itex] with operation * doesn't define a group.
[tex]
\begin{array}{c|cccc}
* & a & b & c & d \\ \hline
a & c & d & a & b \\
b & d & c & b & a \\
c & a & b & c & d \\
d & b & a & d & c \\
\end{array}
[/tex]
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 [tex]b * (d * a) = b * b = b[/tex] and this is what I don't understand. Why is the result of this operation not c?. From the table is says that [tex]b * b = c.[/tex] Any help please thanks.
I'm having trouble understanding why the follow composition table for the set [itex]\left\{ a, b, c, d \right\} [/itex] with operation * doesn't define a group.
[tex]
\begin{array}{c|cccc}
* & a & b & c & d \\ \hline
a & c & d & a & b \\
b & d & c & b & a \\
c & a & b & c & d \\
d & b & a & d & c \\
\end{array}
[/tex]
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 [tex]b * (d * a) = b * b = b[/tex] and this is what I don't understand. Why is the result of this operation not c?. From the table is says that [tex]b * b = c.[/tex] Any help please thanks.