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The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group
Statement
For every finite group , is nilpotent and normal, and every normal nilpotent subgroup of is contained in . See The Fitting subgroup of a finite group.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a finite group , the Fitting subgroup is the product of its -cores (def-p-core-of-a-finite-group). The factors are normal, so their finite product is a normal subgroup and does not depend on the order of multiplication. For the trivial group the product is empty and equals . (The Fitting subgroup of a finite group).
For a finite group , the following are equivalent: is nilpotent; every Sylow subgroup is normal; and is the internal direct product of its Sylow subgroups. (A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product).
Let . They form an internal direct product of if and only if every has a unique expression with , equivalently the multiplication map is an isomorphism. (Internal direct products are external direct products, equivalently every element has a unique factorisation).
If is characteristic in and , then . (If is characteristic in and is normal in , then is normal in ).
The -core is the unique largest normal -subgroup of the finite group . (The -core as the largest normal -subgroup).
Every -subgroup of a finite group is contained in a Sylow -subgroup, and all Sylow -subgroups are conjugate. (Sylow II: in a finite group every -subgroup lies in a conjugate of any Sylow -subgroup, and the Sylow -subgroups form a single conjugacy class).
The order of a subgroup of a finite group divides the order of the group. (Lagrange's theorem: for every subgroup of a finite group ).
Proof
For distinct primes and , normality puts every commutator of an element of with an element of in . By [L7], the order of this intersection divides powers of both and , so the intersection is trivial and the two -cores centralize one another. If a product with equals , then for each the element lies both in the -group and in the commuting product of the other prime-power groups; its order divides two coprime numbers and is therefore . Thus product expressions are unique, so [L3] identifies with the internal direct product of the -cores. The direct-product order shows that is the Sylow -subgroup of ; [L2] makes nilpotent, and [L1] gives its normality in .
If is nilpotent, [L2] makes each Sylow subgroup normal in , and conjugacy [L6] makes it unique. Automorphisms preserve orders, so this unique Sylow subgroup is characteristic in ; [L4] makes it normal in , and the maximality clause [L5] puts it in the corresponding -core.
By [L2], is the product of its Sylow subgroups, and step 2.1 puts every factor in the corresponding -core. Hence .
For the family of -cores is empty and , which is nilpotent and contains every normal nilpotent subgroup. This proves the stated claim.
Depends on
- The Fitting subgroup $F(G)=\prod_p O_p(G)$ of a finite group
- The $p$-core $O_p(G)$ as the largest normal $p$-subgroup
- A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product
- Internal direct products are external direct products, equivalently every element has a unique factorisation
- Sylow II: in a finite group every $p$-subgroup lies in a conjugate of any Sylow $p$-subgroup, and the Sylow $p$-subgroups form a single conjugacy class
- If $K$ is characteristic in $N$ and $N$ is normal in $G$, then $K$ is normal in $G$
- Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 103 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- David A. Craven, Finite Group Theory, Sections 1.4 and 2.3 (standard reference, not scraped)