Latin square property sufficient? So I know that for any group table, Every row must contain distinct group elements and the same holds for every column for a group table. And this property is called the Latin Square property. However, Every time I read a book about Abstract Algebra, They say that the latin square property is necessarily but not sufficient for a table to form a group. However, I have not seen any counter example for that.
So I was wondering if there exists a counter example for a table that has the Latin square property but is not a group ? And what property will it violates from the 4 group axioms ?
I mean I know for sure that closure is not violated.
When I was thinking of examples, The easiest I could think of is to construct a table which has no identity element as follows 
$$
\begin{array}{c|lcr}
 & e & a & b \\
\hline
e & e & b & a \\
a & a & e & b \\
b & b & a & e
\end{array}$$
Is that a valid counter example ?

Taken from http://science.kennesaw.edu/~sellerme/sfehtml/classes/math4361/chapter4section1outline.pdf
Let $G=\{1,2,3,4,5\}$ with multiplication table
\begin{array}{|c||c|c|c|c|c|}
\hline 
*&1&2&3&4&5\\
\hline
\hline
1&1&2&3&4&5\\
\hline
2&2&1&4&5&3\\
\hline
3&3&4&5&2&1\\
\hline
4&4&5&1&3&2\\
\hline
5&5&3&2&1&4\\
\hline
\end{array}
It is easy to see that the bottom right $5\times5$ array is a Latin square. However, we have 
$$
2*(3*4)=2*2=1
$$
and 
$$
(2*3)*4=4*4=3
$$
So this is an example where the associative property is not met. 
