I'm working my way through the second and third sylow theorems in my book. Here's the relevant bit:
We have a group $G$ of order $p^\alpha m$ where $p$ does not divide $m$. We have that $Q$ is a p-subgroup of $G$. From the first sylow theorem, we get that there is a p-subgroup $P$ of $G$. We let $S = \{P_1,P_2,...,P_r\}$ be the set of all conjugates of $P$. We let $G$ act on $S$ by conjugation. Now, we can split up $S$ into a union of disjoint orbits, i.e. $S=O_1 \cup O_2 \cup ... \cup O_s$. We notice that $r = |O_1| + |O_2| + ... + |O_s|$ since $S$ is equal to the union of the orbits. We renumber the elements of $S$ so that the first $s$ elements of $S$ are representatives of the Q-orbits: $P_i \in O_i, 1 \leq i \leq s$ We then see that $|O_i| = |Q: P_i \cap Q|\hspace{5mm} 1 \leq i \leq s$ through some probably irrelevant logic. We now choose $Q=P_1$ and get $|O_1| = |P_1 : P_1 \cap P_1| = 1$
Here's the part where I'm confused: My book states that for all $i > 1$, $P_1 \neq P_i$ implies that $P_1 \cap P_i < P_1$ where $A<B$ denotes that $A$ is a subgroup of $B$ but $A \neq B$. I understand that the intersection of two subgroups is in fact a subgroup. I don't understand however, in this context, why this implies that $P_i$ is not a superset of $P_1$. It then goes on to say that the first part of sylow's theorem implies that $|P_1 : P_1 \cap P_i|>1$.
I'm sorry that there's so much machinery here. The real heart of this question is definitely, what are we allowed to say about the order of conjugate subgroups? Thanks in Advance.
The groups $P_i$ are conjugates of each other, so they are all finite groups of the same order. So one of them cannot strictly contain another one. Since $P_i$ is neither equal to nor properly contains $P_1$, at least one element of $P_1$ does not belong to $P_i$, and their intersection must be a strict subset of $P_1$ (and also of $P_i$). It is of course a subgroup as well. This means also that the index of that intersection in $P_1$ is strictly larger than $1$.