Sylow and TI implies SCDIN

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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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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).

Related facts

Similar facts

Opposite facts

Facts used

  1. Alperin's fusion theorem in terms of tame intersections
  2. Sylow implies order-conjugate


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_is, is in N_G(P), and A and B are conjugate via an element in N_G(P).