$1-2.$ I understand these proofs on pp. 5-6 for Cayley tables but what are the intuitions for
Sudoku property : Every element of the group appears only once in each row and each column.
Symmetric identity property: The identity is in row $r$ and column $c$ $\iff$ It is also in row $c$ and column $r$.
$3.$ Give a table for a binary operation on the set $\{e, a, b\}$ of three elements satisfying the group axioms of existence of identity and inverse but not the associativity axiom.
How can we see associativity fails if one row or column contains the same element $\ge 2$ times? Wikipedia says:
Unfortunately, it is not generally possible to determine whether or not an operation is associative simply by glancing at its Cayley table, as it is with commutativity. This is because associativity depends on a 3 term equation, $(ab)c=a(bc)$, while the Cayley table shows 2-term products. However, Light's associativity test can determine associativity with less effort than brute force.
$4.$ What happened to the 4th group axiom on closure of the operation? Did the question shirk it?
Best Answer
(1-2) The Sudoku property (a.k.a. Latin square property) of group tables follows from the group axioms, so I don't think there is any intuition to be gleaned. As for the symmetric identity property, this follows from commutativity of inverses. Say $g_r$ and $g_c$ are group elements appearing in row $r$ and column $c$, respectively. To say that the identity $e$ appears in the $r^{th}$ row and $c^{th}$ column means that $g_r g_c = e$. But note that we have $g_c g_r = e$ as well. Thus, $e$ also appears on the $c^{th}$ row and $r^{th}$ column.
(3) As far as I know, associativity cannot easily be seen from a group table in general. The question is only asking for a group table from a set of $3$ elements, which is why it is feasible to construct such a group table being asked for. Did you read Light's associativity test?
(4) The question indeed doesn't mention closure, but I think that is meant to be implied. This is just a simple matter of writing some group element in each row / column.
EDIT: We know that $a \neq b \neq e$, otherwise the group would not have order $3$. Note that the table is symmetric about the diagonal except for $ab \neq ba$. We want to use this to show that something goes wrong (i.e. associativity fails). Since $a = ba$, we have that $\begin{align} a\color{magenta}b & = (ba)\color{magenta}b \\ b & = \end{align}$.
Also, $b = ab$ implies $\begin{align} \color{darkorange}{b}b & = \color{darkorange}{b}(ab) \\ e & =\end{align}$. But $\color{darkorange}{b}(ab) = e \neq b = (ba)\color{magenta}b$, which shows that associativity fails.