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Questions on group theory which concern finite groups.

1 vote
1 answer
296 views

The property of self-normalizing subgroup

$G$ is a finite solvable group. Let $\{P_{1}, P_{2}, \dotsc , P_{s}\}$ be a Sylow basis of $G$. We have that $G=P_{1}P_{2}\dotsm P_{s}$. Set \begin{equation} \begin{aligned} %% The alignment is ne …
Moomoo Angel line's user avatar
2 votes
0 answers
160 views

The maximal subgroups of a finite solvable group

$\DeclareMathOperator\Syl{Syl}$$G$ is a finite solvable group and $G=Q\rtimes(P\times R)$, where $P\in \Syl_{p}(G)$ with $P$ is cyclic, $Q\in \Syl_{q}(G)$ with $Q$ is normal elementary abelian, $R\in …
Moomoo Angel line's user avatar
2 votes
1 answer
129 views

The property of subgroups of a finite solvable group

$\DeclareMathOperator\Syl{Syl}$$G$ is a finite solvable group and $G=PQR$, where $P\in \Syl_{p}(G)$, $Q\in \Syl_{q}(G)$, $R\in \Syl_{2}(G)$ and $|R|=2$. Suppose that $C_P(R)=P$ and $C_Q(R)=1$. Since …
Moomoo Angel line's user avatar
5 votes
2 answers
258 views

The property of conjugate subgroup

$\DeclareMathOperator\Syl{Syl}$$G$ is a finite group. $N\unlhd G$, $P\in \Syl_{p}(G)$, and $M$ is a proper subgroup of $P$. Suppose that for any $h\in N_G(P)$, if $MM^h=M^hM$, then $M=M^h$. For any $g …
Moomoo Angel line's user avatar