Glauberman type not implies p-constrained

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This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group of Glauberman type for a prime) need not satisfy the second group property (i.e., p-constrained group)
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Statement

It is possible to have a prime number p and a finite group G such that G is a group of Glauberman type for p but is not a p-constrained group.

Proof

Example of the special linear group

Further information: special linear group:SL(2,5), subgroup structure of special linear group:SL(2,5)

Let G = SL(2,5) and p = 2. Then, O_{p',p}(G) = Z(G) is cyclic of order two, and G/Z(G) is a simple group isomorphic to alternating group:A5. If P is a 2-Sylow subgroup of G, then P is isomorphic to a quaternion group and P \cap O_{p',p}(G) = Z(G), so C_G(P \cap O_{p',p}(G)) = G \not \subseteq O_{p',p}(G). Thus, G is not p-constrained.

On the other hand, Z(J(P)) = Z(G), so O_{p'}(G)N_G(Z(J(P))) = N_G(Z(G)) = G, so G is of Glauberman type with respect to the prime p = 2.