Question is :
Let $G$ be a finiote group of order n = $p^{\alpha}$m, where $p$ is a prime number and if $p^r$ | m but $p^{r+1} \nmid$ m
Let $\mathcal M$ be the set of all subset of $G$ which have $p^{\alpha}$ elements . Then $\mathcal M$ has elements $\left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right) $. Given $M_1,M_2$ $\in \mathcal M$, define $M_1 \thicksim M_2$ if there exist g $\in G$ such that $M_1 = M_2$g. This is an equivalence relation. there is at least one equivalence class of elements in $\mathcal M$ such that number of elements in this class is not multiple of $p^{r+1}$, for if $p^{r+1}$ is a divisor of the size of each equivalence class , then $p^{r+1}$ would be a divisor of of the number of elements in $\mathcal M$. Since $\mathcal M$ has $\left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right) $. and $\ \ p^{r+1} \nmid \left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right) $. , this cannot be the case.
Show that the number of elements in this equivalence class is n ?