Answer to Question #139235 in Abstract Algebra for J

Question

construct cayley table (Z subscript (9) , circle times )and verify its an abelian group.

Answer 1

$\begin{matrix} |& 0 & 1 & 2 &3 &4 &5 &6 & 7& 8\\ \hline 0| & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1|& 1 & 2 & 3 & 4& 5 &6 & 7 & 8 & 0\\ 2 | & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1\\ 3| &3 & 4 & 5 & 6 & 7 & 8 & 0 &1 & 2\\ 4 |& 4 & 5 & 6 & 7 & 8 & 0 &1 & 2 & 3\\ 5|&5 & 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4\\ 6 |& 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4 & 5\\ 7|&7 &8 & 0 & 1 & 2 & 3 & 4 & 5 & 6\\ 8 |& 8 &0 &1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \end{matrix}$

Cayley table of (Z$_{9},+)$ .

The table is symmetric. Hence one can check a+b=b+a for all a,b$\in Z_{9}.$

All the entries in the cayley table are from $0,1,\cdots,8$. Hence closure follows.

The column and row of 0 shows that 0 is the identity.

Every row and column contains 0. Hence inverse exist.

Associativity hold true follows from the fact that addition modulo 9 (any n>0) is an associative operation. Associativity also follows by a laborious check from the cayley table.

Therefore (Z$_{9},+)$ is an abelian group.