# 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)
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## Statement

Suppose $G$ is a finite group, $p$ is a prime, and $P$ is a $p$-Sylow subgroup of $G$, such that $P$ is a TI-subgroup (?) of $G$: any conjugate of $P$ is either equal to $P$ or intersects $P$ trivially. 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 $N_G(P)$.

## Proof

CONVENTION WARNING: This article/section uses the right-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.

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 $g = g_1g_2 \dots g_n$ and $A^{g_1g_2 \dots g_r} \subseteq P \cap Q_{r+1}$ for $1 \le r \le n - 1$.

However, since $P$ is TI, its intersection with any other Sylow subgroup (note also fact (2)) is trivial. Since $A^{g_1g_2\dots g_r} \subseteq P \cap Q_{r+1}$, 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 $N_G(P)$.