Constrained for a prime divisor implies not simple non-abelian

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Revision as of 23:47, 2 March 2009 by Vipul (talk | contribs) (New page: ==Statement== Suppose <math>G</math> is a finite group and <math>p</math> is prime divisor of the order of <math>G</math>. Then, if <math>G</math> is a p-constrained group, <math>...)
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Statement

Suppose G is a finite group and p is prime divisor of the order of G. Then, if G is a p-constrained group, G cannot be a simple non-abelian group.

Definitions used

p-constrained group

Further information: P-constrained group (?)

Suppose G is a finite group and P is a p-Sylow subgroup. We say that G is p-constrained if we have:

C_G(P) \le O_{p',p}(G).

Facts used

  1. Prime power order implies not centerless

Proof

Given: A finite group G, a prime divisor p of the order of G. G is p-constrained.

To prove: G is not a simple non-abelian group.

Proof: Since p divides the order of G, there exists a nontrivial p-Sylow subgroup P of G. By fact (1), Z(P) is nontrivial, so the centralizer C_G(P), which contains Z(P) is nontrivial. Thus, by the definition of p-constrained, we have that O_{p',p}(G) is nontrivial. Thus, it is a nontrivial normal subgroup of G, so we have O_{p',p}(G) = G.

Consider N = O_{p'}(G). If N is trivial, then G = O_{p',p}(G)/O_{p'}(G) is a p-group, and hence by fact (1), Z(G) is a nontrivial normal subgroup, forcing Z(G) = G, forcing G to be abelian.

On the other hand, if O_{p'}(G) = G, then p does not divide the order of G, a contradiction.