| Version 6 |
Version 5 |
| \PMlinkescapeword{factors} |
\PMlinkescapeword{factors} |
| \PMlinkescapeword{independent} |
\PMlinkescapeword{independent} |
| \PMlinkescapeword{refinement} |
\PMlinkescapeword{refinement} |
| \PMlinkescapeword{satisfy} |
\PMlinkescapeword{satisfy} |
| \PMlinkescapeword{theorem} |
\PMlinkescapeword{theorem} |
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| A group $G$ is said to be \emph{polycyclic} if it has a subnormal series |
A group $G$ is said to be \emph{polycyclic} if it has a subnormal series |
| \[\{1\}=G_0\normal G_1\normal\dots\normal G_{n-1}\normal G_n=G\] |
\[\{1\}=G_0\normal G_1\normal\dots\normal G_{n-1}\normal G_n=G\] |
| such that $G_{i+1}/G_i$ is cyclic for each $i=0,\dots,n-1$. |
such that $G_{i+1}/G_i$ is cyclic for each $i=0,\dots,n-1$. |
| (Note that this differs from the definition of a supersolvable group in that it does not require each $G_i$ to be normal in $G$.) |
(Note that this differs from the definition of a supersolvable group in that it does not require each $G_i$ to be normal in $G$.) |
| A subnormal series of this form is called a \emph{polycyclic series}. |
A subnormal series of this form is called a \emph{polycyclic series}. |
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| Polycyclic groups are obviously solvable. |
Polycyclic groups are obviously solvable. |
| In fact, the polycyclic groups are precisely those solvable groups that satisfy the maximal condition (that is, those solvable groups all of whose \PMlinkname{subgroups}{Subgroup} are finitely generated). |
In fact, the polycyclic groups are precisely those solvable groups that satisfy the maximal condition (that is, those solvable groups all of whose \PMlinkname{subgroups}{Subgroup} are finitely generated). |
| In particular, a finite group is polycyclic if and only if it is solvable. |
In particular, a finite group is polycyclic if and only if it is solvable. |
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The \emph{Hirsch number} (or \emph{Hirsch length}) of a polycyclic group $G$ is the number of infinite factors in a polycyclic series of $G$.
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The \emph{Hirsch number} of a polycyclic group $G$ is the number of infinite factors in a polycyclic series of $G$.
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| This is independent of the choice of polycyclic series, as a consequence of the Schreier Refinement Theorem. |
This is independent of the choice of polycyclic series, as a consequence of the Schreier Refinement Theorem. |
