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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

 

Relative normal complements in finite groups


Author: Pamela A. Ferguson
Journal: Proc. Amer. Math. Soc. 87 (1983), 38-40
MSC: Primary 20D10
MathSciNet review: 677226
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Abstract: $ (G,H,{H_0},\pi )$ denotes the following configuration: $ H$ and $ {H_0}$ are the subgroups of the finite group $ G$ with $ {H_0} \trianglelefteq H $ is the set of primes dividing $ (H:{H_0})$. For $ (G,H,{H_0},\pi )$ we consider conditions $ ({\text{A}})$, $ ({{\text{B}}_0})$, and $ ({\text{C}})$: $ ({\text{A}})$ Any two $ \pi $-elements of $ H - {H_0}$ which are $ G$-conjugate are $ H$-conjugate. $ ({{\text{B}}_0})$ For each $ \pi $-element $ x \in H - {H_0}$, $ {C_G}(x) = I(x){C_H}(x)$ where $ I(x)$ is a normal $ \pi '$-subgroup of $ {C_G}(x)$. $ ({\text{C}})\left\vert {{{(H - {H_0})}^{G,\pi }}} \right\vert = (G:H)\left\vert {H - {H_0}} \right\vert$. We show that if $ (G,H,{H_0},\pi )$ satisfies $ ({{\text{B}}_0})$ and $ ({\text{C}})$, or $ ({\text{A}})$ and $ ({{\text{B}}_0})$, and if $ H/{H_0}$ is solvable, then there is a unique relative normal complement $ {G_0}$ of $ H$ over $ {H_0}$.


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DOI: http://dx.doi.org/10.1090/S0002-9939-1983-0677226-5
PII: S 0002-9939(1983)0677226-5
Article copyright: © Copyright 1983 American Mathematical Society