Abelian subgroup of $S_n$ such that $g(a)=b$ for all $a,b \in [1,n]$ 
Let $G$ be an abelian subgroup of $S_n$. Suppose that for all $a,b \in [1,n]$ there is $g$ in $G$ such that $g(a)=b$.
Show that the order of $G$ is $n$.

I don't think that Lagrange's theorem is needed here, unfortunately I don't see what the hypothesis gives us.. Any hint please?

Hint: Let $G$ act on $\{1,\ldots, n\}$ by permutation. Your condition shows that this action is transitive - i.e. the orbit of any element is the whole set. And if $a \in \{1,\ldots, n\}$, $g \in G$ and $g(a) = a$, then $\forall h \in G$, $gh(a) = hg(a) = h(a)$ - so if $g$ stabilises an element $a$, then $g$ stabilises every $b\in\{h(a) : h \in G\}$.
Now use the orbit-stabiliser theorem.
