# 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 $G$ is a finite supersolvable group: a finite group that is also supersolvable. Then, if $d$ is a positive divisor of the order of $G$, $G$ has a subgroup of order $d$.

## Related facts

### 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

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

## Proof

Given: A finite supersolvable group $G$ of order $n$.

To prove: For any positive divisor $d$ of $n$, $G$ has a subgroup of order $d$.

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

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

Thus, $G$ has a subgroup of order $d$ in both cases.