Fill out a group table with 6 elements

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.

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 $\color{red}{2}$.

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.