Sylow and TI implies SCDIN

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow TI-subgroup) must also satisfy the second subgroup property (i.e., SCDIN-subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about Sylow TI-subgroup|Get more facts about SCDIN-subgroup

Statement

Suppose $G$ is a finite group, $p$ is a prime, and $P$ is a $p$ -Sylow subgroup of $G$ . Then $P$ is a SCDIN-subgroup of $G$ : any two elements $A, B$ of $P$ that are conjugate by $g$ in $G$ are conjugate by some element $h$ in $N_{G} (P)$ , and moreover, $h$ and $g$ have the same element-wise action on $A$ .

In fact, something stronger is true: $g$ itself is in $N_{G} (P)$ .

Related facts

Similar facts

Sylow and TI implies CDIN
Sylow implies WNSCDIN: Combines Sylow implies pronormal and pronormal implies WNSCDIN
Sylow implies MWNSCDIN: Combines Sylow implies pronormal and pronormal implies MWNSCDIN
Abelian and Sylow implies SCDIN: Combines Sylow implies pronormal and abelian and pronormal implies SCDIN

Opposite facts

Facts used

Proof

(This proof uses the right-action convention).

Given: A finite group $G$ , a prime $p$ , a $p$ -Sylow subgroup $P$ of $G$ such that the intersection of $P$ with any distinct conjugate of itself is trivial.

To prove: If $A, B \subseteq P$ are conjugate by $g \in G$ , then $A$ and $B$ are conjugate in $N_{G} (P)$ . In fact, $g \in N_{G} (P)$ .

Proof: If $A$ or $B$ is empty or has only one element, the same is true for the other, and in this case we are done. So, we can assume that neither $A$ nor $B$ is contained in the trivial group.

By fact (1), there exist $p$ -Sylow subgroups $Q_{1}, Q_{2}, \dots, Q_{n}$ such that $P \cap Q_{i}$ is a tame intersection, and elements $g_{i} \in N_{G} (P \cap Q_{i})$ such that $g = g_{1} g_{2} \dots g_{n}$ and $A^{g_{1} g_{2} \dots g_{r}} \subseteq P \cap Q_{r + 1}$ for $1 \leq r \leq n - 1$ .

However, since $P$ is TI, its intersection with any other Sylow subgroup (note also fact (2)) is trivial. Since $A^{g_{1} g_{2} \dots g_{r}} \subseteq P \cap Q_{r + 1}$ , we are forced to have that all the $P \cap Q_{i}$ are nontrivial, and hence, equal to $P$ . This forces $Q_{i} = P$ for all $i$ , hence $g_{i} \in N_{G} (P)$ . Hence, $g$ , which is the product of the $g_{i}$ s, is in $N_{G} (P)$ , and $A$ and $B$ are conjugate via an element in $N_{G} (P)$ .