Question about Lemma on Primitive Groups I am reading the following lemma in a paper about graph isomorphism:
Lemma:
Let P be a transitive p-subgroup of Sym(A) with |A| > 1.  Then any minimal p-block system consists of exactly p blocks.  Furthermore, the subgroup P' which stabilizes all of the blocks has index p in P.  
Proof:
The quotient P/P' is a primitive p-group (acting on the blocks) and so the order of P/P' = the number of blocks = p.
Does p-subgroup mean the order is some power of p or p?
I am guessing when they say P' stabilizes all of the blocks, they mean stabilizes all the blocks in a minimal block system, since otherwise one could say each element is a block.  (This paper's definition of a block does not disallow one element blocks).  However, I would think that a minimal block system may not be unique.  Is such a system unique?  Maybe it is.  
Now, I intuitively understand that P/P' is a primitive group acting on the blocks as we are modding out by everything that stabilizes the blocks and so intuitively we are making those many stabilizing permutations the same one thing in the kernel and sort of merging the blocks into a single element in the quotient.  So, intuitively it makes sense to me.  Can someone prove it for me?  Why is the order of P/P' the number of blocks?  That makes no sense to me.  I would assume that the order of P/P' would be (the number of blocks)! since we can permute them as we please.  
 A: I will try and answer some of your questions.
I am going to assume that $A$ is finite, although you didn't say so, because I am not certain that the results carry over to infinite groups.
For a prime $p$, a finite $p$-group is a group has order $p^n$ for some $n \ge 0$.
Yes when it says that $P'$ is the subgroup that stabilizes all blocks, it means all blocks in that particular minimal  system.
There is not necessarily a unique minimal block system, but that does not matter here, because we are working with a specific one.
Let $B$ be the set of blocks in the minimal block system in question. Then the action of $P$ on $A$ induces an action on $B$, so we have a homomorphism $\phi:P \to {\rm Sym}(B)$, of which the kernel is $P'$, so we have $P/P' \cong {\rm im}(\phi) \le {\rm Sym}(B)$.
In the proof, $P/P'$ is being identified with the isomorphic group ${\rm im}(\phi)$, and the minimality of the block system implies that $P/P'$ is acting primitively on $B$.
The main step in the proof is the fact that any primitive $p$-subgroup of ${\rm Sym}(B)$ has order $p$ and acts regularly (or freely if you prefer) on $B$. It follows immediately from that fact that $|B| = |P/P'| = p$.
So why is that true? A permutation group is primitive if and only if it is transitive and its point stabilizer is a maximal subgroup. Now all maximal subgroups of a finite $p$-group have index $p$ and are normal, from which the claim follows.
