Proving that there are only finitely many distinct groups G. Let $n$ be a positive integer. Prove that there are only finitely many distinct groups G on the n-element set {$0,1,2,....,n-1$}, i.e., there are only finitely many Cayley tables for the set {$0,1,2,...,n-1$}. Provide some upper bound, as a function of n, on the number of such group.
I have no idea where to start this problem at all. Can anyone give me any sort of hints/clues to lead me in the right direction?
 A: Cayley's theorem tells you that every group $G$ with $n$ elements is (isomorphic to) a subgroup of $S_n$, which has $n!$ elements. Thus there can be at most $\binom{n!}{n}$ groups on $n$ elements.
Another possible argument is given by looking at the multiplication table of the group. There are $n^{(n^2)}$ ways to fill it, and thus at most that many groups on $n$ elements. For a more precise bound, notice that every line/row must contain every element of the group exactly once (i.e. the multiplication by a fixed element of the group is transitive), so you get easily to the better bound $(n!)^n$.
I am certain that it is possible to find even better bounds, but it probably requires a bit more work.
A: If $|G| = n$, then the group is determined by the product map $f_a(b) = ab$ for each $a$ in the group. For each $a$, the product map $f_a$ gives a permutation of the elements of $G$ (because if $ab = ac$ in a group then $b=c$). So you have at most $n!$ choices for each product map $f_a$, giving a loose bound of $(n!)^n$ on the number of possible groups with elements $\{0,1,\ldots,n-1\}$. Almost certainly, much better bounds are possible.
A: Hint: the Cayley table has $n^2$ squares.  For each square, it must be filled with an element from an $n$-element set.  You can use this to get an upper bound.  It may not be a great upper bound, but you only asked for "some upper bound".
