$(X, \circ)$ is a group, define $a\circ b$ Let $X$ be $\{1,2,3\}$ and $\circ$ a binary operation over $X$ such that $(X,\circ)$ forms a group, and $3\circ 3 = 3$.
Define $a\circ b$ for all $ a,b \in X$. 
I think that the identity element is 3. But I am not sure how to proceed. 
 A: Hints


*

*$3 \circ 3 = 3 \implies 3 \circ 3 \circ 3^{-1} = 3 \circ 3^{-1}
\implies 3 = e$, where $e$  is the identity element / neutral element.

*Remember that the multiplication table is like a sudoku board--every row and column has one of every number $1, 2, 3$.
$$
\begin{array}{|c|ccc|}
\hline
\circ & 1 & 2 & 3 \\
\hline
1 &  &  & 1 \\
2 &  &  & 2 \\
3 & 1 & 2 & 3 \\
\hline
\end{array}
$$

*If $1 \circ 2 = 2$ or $1 \circ 2 = 1$, then $1 = e$ or $2 = e$, which cannot be true.  So $1 \circ 2 = 3$.  Similarly, what is $2 \circ 1$?
A: I'm going to assume that you mean $(X,\circ)$ forms a group. 
With this assumption, we know that $(X,\circ)$ is a group of order three. However, every group of order three is isomorphic to $(\Bbb Z/3\Bbb Z,+)$ (can you prove this?). The Cayley table for $(\Bbb Z/3\Bbb Z,+)$ is
$$
\begin{array}{c|ccc}
+  & 0 & 1 & 2 \\ \hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1
\end{array}
$$
Note that in this Cayley table, the only element $a$ of $\Bbb Z/3\Bbb Z$ that satisfies $a^2=a$ is $a=0$. Thus $3$ is the identity element of $(X,\circ)$, allowing us to fill in part of the Cayley table for $(X,\circ)$:
$$
\begin{array}{c|ccc}
\circ  & 3 & 2 & 1 \\ \hline
3 & 3 & 2 & 1 \\
2 & 2 & ? & ? \\
1 & 1 & ? & ?
\end{array}
$$
Next, note that non-identity elements $a$ in $\Bbb Z/3\Bbb Z$ satisfy $a^2\neq 0$ and $a^2\neq a$. This gives us two more spots in the Cayley table:
$$
\begin{array}{c|ccc}
\circ  & 3 & 2 & 1 \\ \hline
3 & 3 & 2 & 1 \\
2 & 2 & 1 & ? \\
1 & 1 & ? & 2
\end{array}
$$
Finally, note that the two non-identity elements in $\Bbb Z/3\Bbb Z$ are inverses of each other. This completes the Cayley table:
$$
\begin{array}{c|ccc}
\circ  & 3 & 2 & 1 \\ \hline
3 & 3 & 2 & 1 \\
2 & 2 & 1 & 3 \\
1 & 1 & 3 & 2
\end{array}
$$
If you're having trouble proving that every group of order three is isomorphic to $\Bbb Z/3\Bbb Z$, then you can still fill in the Cayley table using @Goos's method.
A: Knowing only $3\circ 3=3$, we start with the table: $$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & & \\ 2 & & & \\ 3 & & & 3\end{array}$$
By definition, a group is closed, associative, has a unique identity element, and every element has an unique inverse.
Closed: $\forall a,b\in X \implies a\circ b \in X$.  Thus the only image of the operator are within the group, $X$.
Associative: $(a\circ b)\circ c = a\circ (b\circ c)$
Identity: $\forall a\in X: a\circ e = e\circ a = a$.  NB: since $3\circ 3 = 3$ then $3$ is the identity element.
Inverse: $\forall a\in X, \exists a^{-1}\in X: a\circ a^{-1} = a^{-1}\circ a = 3$.  The inverse is unique for each element.

Thus we can fill in some more of the graph from the existence of the identity:
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & & 1 \\ 2 & & & 2 \\ 3 & 1 & 2 & 3\end{array}$$
Since the identity is unique $1\circ 1\neq 1$, $2\circ 2 \neq 2$, $1\circ 2\neq 1\neq 2\circ 1$,  $1\circ 2\neq 1\neq 2\circ 1$.  This lets us fill in two more squares.
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & 3 & 1 \\ 2 &3 & & 2 \\ 3 & 1 & 2 & 3\end{array}$$
Noting that these give us the inverse for $1$ and $2$, and since the inverse is unique for each element, then this lets us fill in the final squares: 
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & 2 & 3 & 1 \\ 2 &3 & 1 & 2 \\ 3 & 1 & 2 & 3\end{array}$$
All that remains to do is check that this table is associative.
