Let $G$ be a finite group with $p$ a prime dividing $|G|$. Intersection of two Sylow $p$-subgroups has the max cardinality.

Question

Let $G$ be a finite group with $p$ a prime dividing $|G|$ but not a $p$-group. Suppose that $G$ contains two Sylow $p$-subgroups whose intersection $I$ has maximal cardinality. Show the $N_G(I)$ is not a $p$-group.

$|G|$ means the order of the group $G$ , $N_G(I):=\{x\in G|xIx^{-1}=I\}$.

Let G be a finite group of order of order $p^rm$ with $p$ a prime relatively prime to $m$ and $r > 0$. A subgroup $H$ of $G$ is called a Sylow $p$-subgroup of $G$ if $H$ has order $p^r$.

thanks in advanced.

Answer 1

I will prove something stronger.

Theorem Suppose that $S$ and $T$ are distinct members of $Syl_p(G)$ chosen so that $|S \cap T|$ is maximal among all such intersections. Then the normalizer $N_G(S \cap T)$ has more than one Sylow $p$-subgroup.

Proof Observe that $S \cap T \lt S$: if $S \cap T=S$, then $S=T$, contradicting $S$ and $T$ being distinct. It follows by the "normalizers grow" principle that $S \cap T \subsetneq N_S(S \cap T)$ (and symmetrically $S \cap T \subsetneq N_T(S \cap T)$) and hence the latter is a non-trivial $p$-group. Since $N_S(S \cap T) \subseteq N_G(S \cap T)$, we see that $p$ (and even $|S \cap T| \cdot p$) divides the order of $N_G(S \cap T)$ indeed.

Next, assume that $P \in Syl_p(N_G(S \cap T))$ is its unique Sylow $p$-subgroup. We are going to derive a contradiction. $P$ is a $p$-subgroup of $G$ so it must lie in some Sylow $p$-subgroup of $G$, say $U$. Assume $S \neq U \neq T$ with $P \subseteq U$. Now $N_S(S \cap T)$ is a $p$-subgroup of $N_G(S \cap T)$ and since $P$ is the unique Sylow subgroup of the latter, we must have $S \cap T \subsetneq N_S(S \cap T) \subseteq P \subseteq U$. Hence $S \cap T \subsetneq N_S(S \cap T) \subseteq S \cap U$, violating the maximal choice of $S \cap T$. It follows that $U=S$ or $U=T$.
Suppose for the moment that $P \subseteq S$. Again, $N_T(S \cap T)$ is a $p$-subgroup of $N_G(S \cap T)$ and we get $S \cap T \subsetneq N_T(S \cap T) \subseteq P \subseteq S$, implying $S \cap T \subsetneq N_T(S \cap T) \subseteq S \cap T$, our final contradiction, since a symmetrical argument holds for $P \subseteq T$.