# P-solvable not implies Glauberman type for p

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., p-solvable group) need not satisfy the second group property (i.e., group of Glauberman type for a prime)
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## Statement

It is possible to have a prime number $p$ and a finite group $G$ such that $G$ is $p$-solvable (i.e., is a p-solvable group) but not a group of Glauberman type for $p$.

In fact, this is possible for the primes $p = 2,3$.

## Proof

### Example at the prime two

Further information: symmetric group:S4

Consider the prime $p = 2$ and the group $G = S_4$, the symmetric group on the set $\{ 1,2,3,4 \}$. $G$ is a solvable group, and hence is $2$-solvable. However, $G$ is not of Glauberman type for the prime two, because if we take $P$ as the $2$-Sylow subgroup generated by $\{ (1,2,3,4), (1,3) \}$, then $Z(J(P)) = \{ (), (1,3)(2,4)\}$. In this case, we have:

$O_{2'}(G)N_G(Z(J(P))) = P \ne G$.