$G$ is a group that is transitive on $\{1,2,\dots,n\}$, and let $H_i$ be the subgroup of $G$ that leaves $i$ fixed. Prove that $|G|=n|H_i|$. A subgroup $H$ of the group $S_n$ is called transitive on $B=\{1,2,\dots,n\}$ if for each pair $i,j$ of elements of $B$ there exists an element $h\in H$ such that $h(i)=j$.
Suppose $G$ is a group that is transitive on $\{1,2,\dots,n\}$, and let $H_i$ be the subgroup of $G$ that leaves $i$ fixed: $$H_i=\{g\in G\;|\;g(i)=i\}$$
for $i=1,2,\dots,n$. Prove that $|G|=n|H_i|$.  
Here I observed that:
(i) Each $H_i$ is of same size.
(ii) ${H_i}$ does not form partition of $G$.
What I have done is I define $\phi:H_i\rightarrow H_j$ by $$\phi(g)=hgh^{-1}, \text{where   }h(i)=j,h\in G $$
Such $h$ exists since $G$ is transitive. So it is easy to see that this $\phi$ is bijective. This shows that each $H_i$ is of the same size.

Hint: Use the orbit-stabilizer theorem.
