Completing a Cayley table with few given spaces \begin{array}{ccc}
 * & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\
 \textbf{1} & 1 & & & & & \\ 
\textbf{2} & & & & & 1 & \\
\textbf{3} & & & 1 & & 2 \\
\textbf{4} & & & & & & & \\
\textbf{5} & & & & & & \\
\textbf{6} & & & & & & 1 
\end{array}
need help figuring out the rest of the table

Hint: 1 is the identity because the identity is the only idempotent element in a group.
Edit: (I had some nonsense here before.)
Don't forget that the table for a group can only have symbols appear once in each column and row. Also don't forget that the only orders possible are 1,2,3,6. (You can use this to eliminate the possibility that $5^3=1$, thus this group is cyclic.)
This question: About group multiplication table has helpful tips in the answers, along these lines.

OK, I'm starting to not believe in this problem (if we're in a group). Looking only at the powers of 5, we know $5^2\neq1$ (since $5*2=1$) and we know that $5^3\neq1$ (since $5^3=1\implies 2*5^3=5^2=2$, a contradiction with $3*5=2$) thus $5^6=1$.
But then, multiplying with 2's, $3*5=2 \implies3=2^2$. But then $5^6=1\implies 5^2=2^4\implies 5^2=3^2$, but $3^2=1$, a contradiciton with $5^2\neq 1$. 

TaraB pointed out that $5^6=1$ regardless of the order of $5$, and so you can skip to the final contradiction directly!
