TITLE: Fill out a group table with 6 elements
QUESTION [6 upvotes]: 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?

REPLY [6 votes]: Let's take it slow
$\begin{array}{ c| c | c | c | c |c|c|}
   * & 0& 1 & 2 & 3& 4 & 5\\
   \hline
   0 & 0 & 1  & 2 & 3 & 4 & 5 \\ 
   \hline
   1 & 1 & 2 & 0 & 4 &    &  \\ 
   \hline
   2 &  2 &    &    &   &     & \\ 
   \hline
   3 & 3  & 5  &  &   &  & 1\\ 
   \hline
   4 & 4  &    &  &   &  &  \\ 
   \hline
   5 &  5 &   &  &   & & \\ 
   \hline
  \end{array}$
As each row and column must be distinct we can start by "soduku"ing the $1*a$ row.
$\begin{array}{ c| c | c | c | c |c|c|}
   * & 0& 1 & 2 & 3& 4 & 5\\
   \hline
   0 & 0 & 1  & 2 & 3 & 4 & 5 \\ 
   \hline
   1 & 1 & 2 & 0 & 4 &  \color{blue}5  & \color{blue}3 \\ 
   \hline
   2 &  2 &    &    &   &     & \\ 
   \hline
   3 & 3  & 5  &  &   &  & 1\\ 
   \hline
   4 & 4  &    &  &   &  &  \\ 
   \hline
   5 &  5 &   &  &   & & \\ 
   \hline
  \end{array}$
Now $1*2=0$ so $2=-1$ So if we replace $2$ with $-1$ and use $\pm 1*m=k\iff m=\pm 1* k$ and $k*\pm 1=m \iff k=m*\mp 1$ we can get the following ($\color{green}{green}$ for $1/-1$ and $\color{blue}{blue}$ for "sudoku"ing):
$\begin{array}{ c| c | c | c | c |c|c|}
   * & 0& 1 & -1 & 3& 4 & 5\\
   \hline
   0 & 0 & 1  & -1 & 3 & 4 & 5 \\ 
   \hline
   1 & 1 & -1 & 0 & 4 &  \color{blue}5  & \color{blue}3 \\ 
   \hline
   -1 &  -1 &  \color{green}0  &  \color{green} 1  &  \color{green}5 & \color{green}3    & \color{green}4\\ 
   \hline
   3 & 3  & 5  & \color{green}4 &   &  & 1\\ 
   \hline
   4 & 4  &  \color{blue}3  & \color{green}5 &   &  &  \\ 
   \hline
   5 &  5 & \color{blue}4  & \color{green}3 &   & & \\ 
   \hline
  \end{array}$
Now $3*5 = 1$ so $3*5*-1 = 0$ and $3*3=0$
And knowing $a*b = a*0*b = (a*1)(-1*b)=(a*-1)(1*b)$ we see:  $3*3= 4*1*-1*4=4*4;3*3=5*-1*1*5=5*5=0$.
$\begin{array}{ c| c | c | c | c |c|c|}
   * & 0& 1 & -1 & 3& 4 & 5\\
   \hline
   0 & 0 & 1  & -1 & 3 & 4 & 5 \\ 
   \hline
   1 & 1 & -1 & 0 & 4 &  \color{blue}5  & \color{blue}3 \\ 
   \hline
   -1 &  -1 &  \color{green}0  &  \color{green} 1  &  \color{green}5 & \color{green}3    & \color{green}4\\ 
   \hline
   3 & 3  & 5  & \color{green}4 & \color{red}0  &  & 1\\ 
   \hline
   4 & 4  &  \color{blue}3  & \color{green}5 &   & \color{red}0 &  \\ 
   \hline
   5 &  5 & \color{blue}4  & \color{green}3 &   & & \color{red}0\\ 
   \hline
  \end{array}$
We can "sudoku" the rest (and replace $-1$ with $2$)
$\begin{array}{ c| c | c | c | c |c|c|}
   * & 0& 1 & 2 & 3& 4 & 5\\
   \hline
   0 & 0 & 1  & 2 & 3 & 4 & 5 \\ 
   \hline
   1 & 1 & 2 & 0 & 4 &  \color{blue}5  & \color{blue}3 \\ 
   \hline
   2 &  2 &  \color{green}0  &  \color{green} 1  &  \color{green}5 & \color{green}3    & \color{green}4\\ 
   \hline
   3 & 3  & 5  & \color{green}4 & \color{red}0  & \color{blue}2 & 1\\ 
   \hline
   4 & 4  &  \color{blue}3  & \color{green}5 & \color{blue}1  & \color{red}0 & \color{blue}2 \\ 
   \hline
   5 &  5 & \color{blue}4  & \color{green}3 & \color{blue}2  & \color{blue}1& \color{red}0\\ 
   \hline
  \end{array}$
Hmm, though actually we haven't verified that what we were given would be a group.  We just took it on faith.  But we can verify by associativity it holds.
====
Here's an expansion and David Splutterwit's answer.  This is hopefully not such a haphazard guessing game and hopefully will unquestionably be a group.
First, let's not use numbers, that confusing.
Let's use $e=0,a=1,b=2,c=3,d=4,g=5$.($e$ is the identity.)
We know $a^2 = b$ $(1*1=2)$ and $a*b = a*a^2 = a^3=0$ $(1*2 = 0)$
So we have $\{e,a, a^2, c,d,g\}$
We have the three following relations: $a*c=d$, $c*a=g$, and $c*g=c^2a=a$.  Which would therefore mean $c^2 = e$
So we have $\{e,a,a^2,c,ac,ca|a^3 = c^2= e\}$. $ac\ne ca$. 
Now $a*e = a, a*a= a^2, a*a^2 = e, a*c = ac$ and $a*ac\ne ac$.  So $a*ac = ca$
So we have $\{e, a, a^2, c,ac,a^2c|a^3=c^2=e, a^2c = ca\}$.
And that's enough to describe a group.  
Or $\{0, 1, 1^2=2, 3, 1*3=4, 1^23=5|1^3=3^2=0, 1^23=3*1\}$.
You can fill in the table from there.
Wish I had thought of that first.  It's much more "grown-up" than filling in a table.