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Nilpotence lifts over the Frattini subgroup of a finite group
Statement
Let be finite and let . Then is nilpotent if and only if is nilpotent. In particular, is nilpotent if and only if is nilpotent. See The Frattini subgroup as the intersection of the maximal subgroups of a finite group.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a finite group , the Frattini subgroup is If , the family is empty and its intersection inside is itself. Thus . (The Frattini subgroup as the intersection of the maximal subgroups of a finite group).
If is finite and is a Sylow -subgroup of , then . (Frattini argument: if and is Sylow in , then ).
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).
For , subgroups of correspond to subgroups of containing , and the correspondence preserves normality. (Correspondence theorem: subgroups of correspond to subgroups of containing , with normality preserved).
Every subgroup and every quotient of a nilpotent group is nilpotent, and every finite direct product of nilpotent groups is nilpotent. (Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent).
Let be finite, let be a Sylow -subgroup, and let be a -subgroup. There is with ; in particular 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).
If is characteristic in and , then . (If is characteristic in and is normal in , then is normal in ).
Proof
If is nilpotent, then its quotient is nilpotent by [L5].
Conversely, assume that is nilpotent, and let be a Sylow -subgroup of . Every automorphism of permutes its maximal proper subgroups, so [L1] makes characteristic and hence normal in . If is Sylow in , then [L6] gives with ; normality of gives . Thus contains a Sylow -subgroup of and, being a -subgroup, is itself Sylow there. Consequently is a Sylow -subgroup of .
By nilpotence and [L3], this quotient Sylow subgroup is normal; conjugacy [L6] makes it unique, hence characteristic in . Since , [L7] and [L4] give .
Because is Sylow in , the product-order formula shows that is Sylow in . Apply the Frattini argument [L2] to the normal subgroup : If were proper, finiteness would place it in a maximal subgroup ; [L1] gives , contradicting the displayed equality. Hence .
Thus every Sylow subgroup of is normal, so [L3] makes nilpotent. Together with step 1.1 this proves both directions.
Taking gives the asserted special case. When , one has , and both groups in the equivalence are trivial and nilpotent.
Depends on
- The Frattini subgroup $\Phi(G)$ as the intersection of the maximal subgroups of a finite group
- Frattini argument: if $N\trianglelefteq G$ and $P$ is Sylow in $N$, then $G=N N_G(P)$
- A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product
- Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$, with normality preserved
- Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent
- 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 $G$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 73 results over 14 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)