I have a problem with one of my questions. The question is:
complete the table by providing an example of a binary operation $*$ defined on $\{a , b ,c\}$ such that $*$ is commutative and has the identity element $c$. \begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & ? & ? \\ b & ? & b & ? \\ c & ? & ? & c \\ \end{array}
(I need the letters in place of the ?)
I don't understand what they want me to do ???
$c$ is the identity and the operation is commutative so $$a\cdot c=c\cdot a=a$$ Similarly $$b\cdot c=c\cdot b=b$$ It seems to me that $a\cdot b$ can be $a$ or $b$.
You have to fill out the Cayley table such that the operation is commutative with identity element 'c'. There is more than one way to accomplish that, for example the operation a*b=max{a,b} with (a,b,c) = (2,1,0) will satisfy the preconditions given in the table.
For $a$ to be the identity element, it must be $a\ast b=b=b\ast a$ and $a\ast c=c=c\ast a$, so you have: \begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & \color{lime}{b} & \color{lime}{c}\\ b & \color{lime}{b} & b & ? \\ c & \color{lime}{c} & ? & c \\ \end{array} Then, for commutativity to hold, this is for the Cayley table to be symmetric, you have only three choices:
\begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & b & c\\ b & b & b & \color{red}{a} \\ c & c & \color{red}{a} & c \\ \end{array}
\begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & b & c\\ b & b & b & \color{red}{b} \\ c & c & \color{red}{b} & c \\ \end{array}
\begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & b & c\\ b & b & b & \color{red}{c} \\ c & c & \color{red}{c} & c \\ \end{array}
Exactly at the same way, you have three choices if $b$ is the identity, and three more if the identity is $c$. In this last case:
\begin{array}{c|ccc} \ast & a & b & c \\ \hline a & a & ? & \color{lime}{a}\\ b & ? & b & \color{lime}{b} \\ c & \color{lime}{a} & \color{lime}{b} & c \\ \end{array} and so on, as above.
Let a binary operation $*$ on N be defined b $a*b=HCF$ of $a$ and $b$.
Construct a composition Tabe that the set $H=(1,2,3,4,5,6,)$ is closed under $*$.
