2-Sylow subgroup is TI implies it is normal or there is exactly one conjugacy class of involutions

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Suppose G is a finite group of even order with the property that one (and hence all) of its 2-Sylow subgroups is a TI-subgroup. In other words, any two distinct 2-Sylow subgroups intersect trivially. Then, one of the following is true:

  1. G has exactly one conjugacy class of involutions (i.e., any two elements of order two are conjugate).
  2. The 2-Sylow subgroup of G is normal, or equivalently, G has a unique 2-Sylow subgroup.

Facts used

  1. Sylow implies order-dominating
  2. Involutions are either conjugate or have an involution centralizing both of them


Given: A finite group G of even order whose 2-Sylow subgroups are TI.

To prove: Either G has a single conjugacy class of involutions or G has a normal 2-Sylow subgroup.

Proof: Suppose G has two distinct conjugacy classes of involutions. Let P,Q be any two 2-Sylow subgroups of G. First, note that every involution is conjugate to some element in any given 2-Sylow subgroup (fact (1), the domination part of Sylow's theorem), so every 2-Sylow subgroup intersects every conjugacy class of involutions. Thus, we may pick x \in P, y \in Q as representatives of the two (distinct) conjugacy classes of involutions.

By fact (2), there exists an involution z that centralizes both x and y. Since z is an involution, it is contained in a 2-Sylow subgroup R. Because the 2-Sylow subgroups are TI, R is the unique 2-Sylow subgroup containing z. Similarly, P is the unique 2-Sylow subgroup containing x and Q is the unique 2-Sylow subgroup containing y.

Since x and z are commuting involutions, \langle x,z \rangle is a 2-group. By fact (1), there is a 2-Sylow subgroup containing this. This must equal both P and R, so P = R. Similarly, \langle y,z \rangle is a 2-group, so there is a 2-Sylow subgroup containing it, which must equal both Q and R, so Q = R. Thus, P = Q. Since P,Q were arbitrary, we conclude that any two 2-Sylow subgroups are equal, so there is a unique 2-Sylow subgroup which must therefore be normal.


Textbook references