Finite supersolvable implies subgroups of all orders dividing the group order
This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finite supersolvable group) must also satisfy the second group property (i.e., group having subgroups of all orders dividing the group order)
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- Finite nilpotent implies every normal subgroup contains normal subgroups of all orders dividing its order
- Finite nilpotent implies every normal subgroup is part of a chief series
The converse of this statement is false -- we can have a non-supersolvable group that has subgroups of all orders dividing the group order. In fact, every finite solvable group can be embedded inside a group with the property. See every finite solvable group is a subgroup of a finite group having subgroups of all orders dividing the group order.
- Cyclic implies every subgroup is characteristic
- Characteristic of normal implies normal
- Lagrange's theorem
- Supersolvability is quotient-closed
- Normal Hall implies permutably complemented: This is the existence part of the Schur-Zassenhaus theorem.
We prove this by induction. The base case for induction, the case of a group of order 1, is trivially true. We therefore assume that the resultis proved for all groups of order smalelr than the group we are trying to prove the result for.
Given: A finite supersolvable group of order .
To prove: For any positive divisor of , has a subgroup of order .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||has a cyclic normal subgroup of prime order , for some||Facts (1), (2)||is supersolvable of order||By the definition of supersolvability, has a nontrivial cyclic normal subgroup, say . Let be a prime dividing the order of . has a subgroup of order , which by Fact (1) is characteristic in C. By Fact (2), is normal in .|
|2||has subgroups of every order dividing .||inductive assumption||By Fact (3), has order , which is strictly smaller than . Also, by Fact (4), it is supersolvable. Therefore, the inductive assumption applies, and it has subgroups of all orders dividing it.|
|3||If , then has a subgroup of order||Steps (1), (2)||Since and , we get that . Therefore, has a subgroup of order . By Fact (4), the inverse image of this in has order .|
|4||If does not divide , has a subgroup of order containing as a cyclic normal subgroup of order .||Steps (1), (2)||Since divides but does not divide , . Therefore, has a subgroup of order . By Fact (4), the inverse image of this in has order . Call this group .|
|5||If does not divide , has a subgroup of order .||Step (4)||Continuing from Step (4), is a normal Sylow subgroup of (since does not divide ). Hence, fact (5) tells us that contains a complement to , which must therefore have order .|
|6||In all cases, has a subgroup of order||Steps (3), (5)||The two cases were dealt with in steps (3) and (5).|
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