Trouble with filling out a Cayley table I have trouble with filling this table:
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & & & q & & & \\ 
\textbf{q} & r & & & & & \\
\textbf{r} & & & & & r & \\
\textbf{s} & & p & & & & \\
\textbf{t} & & & & & & \\
\textbf{u} & t & & & & & p 
\end{array}$$
It needs to be filled out so it forms a group. I only figured out that t is the neutral element (because $r ∗ t = r$?). How do I figure out the rest of the table?
Thanks.
 A: $r*t=r$, so you can fill out the rest of the $t$ row and column like that. Since it's the identity, $p*t=t*p=p$, etc.
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & & & q & &p & \\ 
\textbf{q} & r & & & &q & \\
\textbf{r} & & & & & r & \\
\textbf{s} & & p & & & s & \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t & & & &u & p 
\end{array}$$
Then we need to use a certain fact: each element will only appear in any row or column once. Applying this to the bottom row, we see that the only option for $u*?=q$ is when $u*s=q$, since the other columns already contain $q$ (this is like Sudoku). Then $u*r$ can't be $r$, since $u$ isn't the identity. We must have that $u*r=s$. Filling in the bottom row:
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & & & q & &p & \\ 
\textbf{q} & r & & & &q & \\
\textbf{r} & & & & & r & \\
\textbf{s} & & p & & & s & \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t &r & s&q &u & p 
\end{array}$$
Note that $u*p=t$. So $u$ and $p$ are inverses. This implies that $p*u=t$. This lets us fill in the right column the same way we did the bottom row.
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & & & q & &p &t \\ 
\textbf{q} & r & & & &q & s\\
\textbf{r} & & & & & r & q\\
\textbf{s} & & p & & & s &r \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t &r & s&q &u & p 
\end{array}$$
Now, since $p$ and $u$ are inverses, $u*s=q\implies p*u*s=p*q\implies s=p*q$. Applying this analysis, we fill in the top row and left column. 
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & u& s& q &r &p &t \\ 
\textbf{q} & r & & & &q & s\\
\textbf{r} & s& & & & r & q\\
\textbf{s} & q& p & & & s &r \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t &r & s&q &u & p 
\end{array}$$
Now it turns out that $q$ is its own inverse. Because $p*q=s\implies p=s*q^{-1}$ and $s*q=p$. So $q*q=t$
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & u& s& q &r &p &t \\ 
\textbf{q} & r &t & & &q & s\\
\textbf{r} & s& u& & & r & q\\
\textbf{s} & q& p & & & s &r \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t &r & s&q &u & p 
\end{array}$$
$r=q*p\implies q*r=p$, since $q$ is its own inverse. This lets us fill in the rest of the table easily.
$$\begin{array}{ccc}
 * & \textbf{p} & \textbf{q} & \textbf{r} & \textbf{s} & \textbf{t} & \textbf{u} \\
 \textbf{p} & u& s& q &r &p &t \\ 
\textbf{q} & r &t & p& u&q & s\\
\textbf{r} & s& u& t&p & r & q\\
\textbf{s} & q& p & u& t& s &r \\
\textbf{t} & p & q & r & s & t & u \\
\textbf{u} & t &r & s&q &u & p 
\end{array}$$
