Fill out a group table with 6 elements 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?
 A: I am guessing it is isomorphic to the dihedral group on $6 $ elements. You have gone wrong in the last "quater" of the grid.
\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  & \color{red}{0}  & \color{blue}{2} & \color{blue}{1}\\ 
   \hline
   4 & 4  &  3  & 5 & \color{blue}{1}  & \color{red}{0} & \color{blue}{2} \\ 
   \hline
   5 &  5 & 4  & 3 & \color{blue}{2}  & \color{blue}{1}& \color{red}{0}\\ 
   \hline
  \end{array}
$D_6 = \{ e,a,a^2,b,ab,a^2b \mid a^3=b^2=e \, \, \,  ab=ba^2 \}$ The elements $\color{red}{b,ab \text{ and } a^2 b}$ are of order $\color{red}{2}$.
A: 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.
