Let $G $ be group such that $p^a $divides $|G|$ then G has subgroup of order $p^a|$.
Proof:Let $|G|=p^am$
Let $\mu$ is set of all subset with $p^a$ elements.
SO there are $\binom{p^am}{p^a}$ element .We can easily prove There exist some power of such that $p^r$ divides m but $p^{r+1}$ does not divide m.Then
$p^{r+1}$ does not divide $\binom{p^am}{p^a}$
COnstruct Equivalence class in$\mu$ such that $M_1 \sim M_2$ if $\exists g\in G $ such that $M_1g=M_2$ .
There exist atleast one class such that $p^{r+1}$ doesnot divide its size.Other wise if not then $p^{r+1}$ divide $\binom{p^am}{p^a}$
That equivalence class is say {$M_1,M_2,.....,M_n$}
H={$g\in G|M_1g=M_1$} It is easy to show that H is subgroup.
From Here It is to prove that $|G|=n\times |H|$ Which I don't able to write.
(Please help me to prove that)
$p^am=n|H|$ also $p^r/n$ But $p^{r+1}$ doesnot divide n
$p^(a+r)/p^am$ then $p^{a+r}/n|H|$ hence $p^r/|H|$
Hence $p^r \leq |H|$
By defination of H $m_1h\in M_1 \forall h\in H$ hence $M_1$ has atleast |H| elements but $M_1$ has $p^a$ element hence $p^a\geq |H|$ which implies |H|=$p^a$ Hence Done.
I am not able to get argument for |G|=n|H|
Any Help will be appreciated.
Now that I've understood the question here is the answer:
Consider the $G$ action on $\mu$ by right multiplication and consider the element $M_1\in \mu$.
By the orbit stabilizer theorem
$$|Stab(M_1)|\cdot |Orb(M_1)|=|G|$$
But by definition $$Stab(M_1)=\{g\in G|M_1g=M_1\}=H$$ and $$Orb(M_1)=\{M_1g|g\in G\}=\{M_1,\dots,M_n\}$$
Thus
$$|H|\cdot n=|G|.$$
EDIT: I leave the previous answer for completeness
If you need an explanation for the first inequality, just ask.
For the second:
Basically you are using that left group multiplication is injective (as groups have inverses), in the sense that $$\{m_1h_1,m_1h_2,\dots,m_1h_{|H|}\}\subset M$$ are all different. Thus taking cardinalities $$|H|=|\{m_1h_1,m_1h_2,\dots,m_1h_{|H|}\}|\leq |M|=p^a$$