The size of normalizer of intersection of two distinct Sylow $p$-subgroups

Question

Let $G$ be a group.
Let $P\in Syl_p(G)$ for some prime $p$ and $|P|=p^a,a\geq 2$.
Suppose there exists $R\in Syl_p(G)$ with $R\neq P$ and $P\cap R\neq1$.
Let $P_0=P\cap R$.
Assume $|P_0|=p^{a-1}$. Then $P_0\lhd P$ and $P_0 \lhd R$.
So $N_G(P_0)$ contains $P$ and $R$.
In particular, $|N_G(P_0)|=p^ak$, where by Sylow's Theorem, $k\geq p+1$.

I can't get the point that how the Sylow's Theorem make the value of $k$ to be larger than $p+1$. Obviously it can't be $p$ otherwise it contradicts the definition of Sylow's $p$-subgroup. But what will be happened if $1\leq k

Answer 1

You are expected to apply Sylow's theorems to the group $N_G(P_0)$.

How many Sylow $p$-subgroups do you think it has? Let $n_p$ be their number.

• $R$ and $P$ are there so $n_p\ge2$.
• By Sylow's theorems $n_p\equiv1\pmod p$.
• Also by Sylow's theorems $n_p\mid k$.

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[Edit after the OP figured it out:] With these data points at hand we can conclude that as $n_p>1$ we must have $$1+p\le n_p\le k.$$ [/Edit]