<s>[INST] $(X, \circ)$ 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. 
 [/INST] 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.
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