Does this composition table necessarily define a group? We know that each row (and each column) of composition table of a finite group, is a rearrangement (permutation) of the elements of the group.
How about the other way round? If we have a composition table where each row and each column is a permutation of the elements of a set, does this composition table necessarily define a group?
If not then give a counter example.
 A: What you describe is a quasigroup.
A quasigroup is an ordered pair $(A,\cdot)$, where $A$ is a set, and $\cdot$ is a binary operation on $A$ with the property that for all $a,b\in A$ there exist unique solutions to the equations $a\cdot x = b$ and $y\cdot a=b$. If you think in terms of the Cayley table, you ask that each row and each column contain each element of $A$ exactly once; that is, that the Cayley table be a Latin square.
Quasigroups that are not groups exist for all orders greater than or equal to $3$; if you allow the empty set, it is also a quasigroup that is not a group.
A quasigroup is a group if and only if the operation is associative.
A: No.
Consider
$$\begin{array}{c|ccc}
\ast & e & a & b\\
\hline
e & e & a & b\\
a & b & e & a\\
b & a & b & e
\end{array}.$$
There is no identity, so it cannot be a group.
A: For an example with an identity element and inverses, consider
$$\begin{array}{c|ccccc}
\ast & e & a & b & c & d\\
\hline
e & e & a & b & c & d \\
a & a & e & c & d & b \\
b & b & d & e & a & c \\
c & c & b & d & e & a \\
d & d & c & a & b & e
\end{array}.$$
It is easy to find triples that do not associate, but the nonassociativity can also be seen from the fact that this loop (quasigroup with identity element) has order 5 but every element has order 2, hence Lagrange's theorem does not hold.
