# 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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## Statement

Suppose is a finite supersolvable group: a finite group that is also supersolvable. Then, if is a positive divisor of the order of , has a subgroup of order .

## Related facts

### Similar facts

- 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

### Converse

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.

The smallest examples of non-supersolvable groups with subgroups of all orders dividing their order are symmetric group:S4 and direct product of A4 and Z2, both of order 24.

## Facts used

- Supersolvability is subgroup-closed
- Supersolvability is quotient-closed
- Characteristic of normal implies normal
- Normal Hall implies permutably complemented: This is the
*existence*part of the Schur-Zassenhaus theorem.

## Proof

**Given**: A finite supersolvable group of order .

**To prove**: For any positive divisor of , has a subgroup of order .

**Proof**: We prove this by induction on the order of . Note that by facts (1) and (2), every proper subgroup and every proper quotient of is supersolvable, and hence, by the inductive assumption, has subgroups of all orders dividing its order.

- has a cyclic normal subgroup of prime order , for some : By the definition of supersolvability, has a nontrivial cyclic normal subgroup, say . , being cyclic, has characteristic subgroups of all orders dividing the order of . In particular, has a cyclic subgroup of order that is characteristic in it. By fact (3), this subgroup is normal in , so we get a normal subgroup of order in .
- If , then has a subgroup of order : has a subgroup of order , and the inverse image of this via the quotient map is a subgroup of order in .
- If does not divide , then by the previous step, still has a subgroup of order (it may be that ).
- Continuing from the previous step, contains a subgroup of order . In particular, has a subgroup of order : is a normal Sylow subgroup of (since does not divide ). Hence, fact (4) tells us that contains a complement to , which must therefore have order .

Thus, has a subgroup of order in both cases.