Group acting on its subsets Let $ G $ be a group with $ |G|=mp^\alpha $ where $ \alpha\geq1 $ and p is prime integer with $p \nmid m$. Then denote the set of subsets of G, having $p^\alpha$ size, with $X$.
Then with the action $g \cdot S=\lbrace gs \mid s \in S\rbrace$ for $S \in X$
show that the stabilizer $G_S$ is a p-subgroup of $G$.
Any ideas for how to approach to this question will be appreciated.
 A: I was going at it all wrong, the idea is to prove all of the elements of the stabilizer have orders which are powers of $p$.
To see this consider an element $g$ of the stabilizer of $S$, it defines a permutation on the set $S$ in which all of the cycles have length equal to the order of $g$. Then the length of the cycles needs to be a divisor of the number of elements in $S$, which is $p^\alpha$. So the order of $g$ is a divisor of $p^\alpha$, hence a prime power.
Since the stabilizer has only elements of order powers of $p$ the stabilizer has prime power order.
Note more generally this strategy works if you are working on subsets of size $k$, where $\gcd(|G|,k)$ is a prime power.
A: The cardinality of your set is $$N=\binom{p^\alpha m}{p^\alpha}$$
You should show that $(p,m)=1$ implies $N$ is not divisible by $p$. Now, we know that if $G$ acts on the set $S$ of subsets of size $p^\alpha$, the cardinality of $N$ of $S$ equals the sum of the orbits of this action. That is, $$N=\sum_{i=1}^n | Gx_i|$$ where each $x_i$ is a representative of a unique orbit of the $G$-action. Since $p$ doesn't divide $N$, we know that $p$ doesn't divide some of the terms $|Gx_i|$. If $p$ doesn't divide $|Gx_i|$, from $|Gx_i|=|G:{\rm stab}\; x_i|$ it follows that the stabilizer ${\rm stab}\; x_i$ is such that its cardinality $M=p^\alpha n$, $n\mid m$. Can you show that indeed $n=1$? 
Note that what you're showing is that some stabilizer is a $p$-group. 
A: Hint: Take two $x,y\in G_S$ and then try to find that $xy^{-1}\in G_S$. Completing this scheme assures you that $G_S<G$. 
Now, since in one hand we have $$\#X={p^{\alpha}m \choose p^{\alpha}},$$ 
and in another
$$\#X= \sum_{S_i}\#{\rm Orb}(S_i)=\sum_{S_i}[G:G_{S_i}],$$ 
but $p\nmid {p^{\alpha}m \choose p^{\alpha}}$,  then there exists one $S_0\in X$ such that $p\nmid [G:G_{S_0}]$.
This implies that $p$ divides $|G_{S_0}|$. This is because if $p\nmid[G:G_{S_0}]$ then they are coprime, meaning that exists $\lambda,\mu$ integers such that $\lambda p+\mu[G:G_{S_0}]=1$, so $|G_{S_0}|=\lambda p|G_{S_0}|+\mu p^{\alpha}m$ hence the claim. 
Similarly also $p^2$ and $p^3$, etcetera, will do divide $|G_{S_0}|$. 
Clearly we can proceed just to $p^{\alpha}$ dividing $|G_{S_0}|$.
So $|G_{S_0}|=p^{\alpha}n$ for some $n\in{\Bbb{N}}$. 
At this point maybe you can complete de proof. I still can't. As soon I discover, I'll tell you.
