# Intuition for Cayley Table and Cayley Table for identity, inverse but not associativity - Fraleigh p. 47 4.24

$$1-2.$$ 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?

(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.

• Thanks. I upvoted. For 1, is there some better way to see the Sudoku property directly? The linked proof is by contradction. Hence it doesn't help. For 3, how did they prognosticate (please see my profile) this table? I can't see how they fabricated it? I didn't look at that test because it's not in my book. Hence I don't think we need it. – Group Theory Dec 24 '13 at 7:43
• See the edit for $3$. As for $1$, I'm not sure what you are asking. The table comes from the axioms of a group. – tylerc0816 Dec 24 '13 at 16:09
• For 1, I was trying to explain because the linked proof is by contradiction, it doesn't help me to prognosticate (please see profile) the Sudoku property. I am trying to understand how you can see the Suduko property before proving anything or fabricating a counterexample. Is this clearer? – Group Theory Dec 25 '13 at 5:03
• Thanks for the edit. For 3, I still don't see how you'd prognosticate this table. The starting point is to 'make some row or column contain some element twice' so 'it can't be a group.' Where does this fact come from? – Group Theory Dec 25 '13 at 5:04