Let $G=\{0,1,2,3,4,5\}$ be a group whose table is partially shown below:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & & \\
\hline
2 & 2 & & & & & \\
\hline
3 & 3 & 5 & & & & 1\\
\hline
4 & 4 & & & & & \\
\hline
5 & 5 & & & & & \\
\hline
\end{array}
Complete the table.
Needed to use Inverses, cancellation and No 1 element can repeat per row/col, but got the wrong table.
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & 5 & 3 \\
\hline
2 & 2 & 0 & 1 & 5 & 3 & 4 \\
\hline
3 & 3 & 5 & 4 & 2 & 0 & 1\\
\hline
4 & 4 & 3 & 5 & 0 & 1 & 2 \\
\hline
5 & 5 & 4 & 3 & 1 & 2& 0\\
\hline
\end{array}
What is the correct table?
Best Answer
Suppose you had got this far and needed to fill out the rest:
\begin{array}{ c| c | c | c | c |c|c|} * & 0& 1 & 2 & 3& 4 & 5\\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 0 & 4 & 5 & 3 \\ \hline 2 & 2 & 0 & 1 & 5 & 3 & 4 \\ \hline 3 & 3 & 5 & 4 & ? \\ \hline 4 & 4 & 3 & 5 \\ \hline 5 & 5 & 4 & 3 \\ \hline \end{array}
One way to proceed would be to use the property of associativity, which says that $(ab)c = a(bc)$. Let $a = 3^2$, and consider $3^3$: since $(3^2)3 = 3(3^2)$, we have $a\cdot3=3\cdot a$. In other words, $3$ and $a$ commute, and since $a \in \{0,1,2\}$, we see that $a=0$.
It's important to understand that the group multiplication table has more properties than just containing a permutation of the elements in each row and column; otherwise you merely have a Latin square.