# Finite supersolvable implies subgroups of all orders dividing the group order

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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. Cyclic implies every subgroup is characteristic
2. Characteristic of normal implies normal
3. Lagrange's theorem
4. Supersolvability is quotient-closed
5. Normal Hall implies permutably complemented: This is the existence part of the Schur-Zassenhaus theorem.

## Proof

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 $G$ of order $n$.

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

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $G$ has a cyclic normal subgroup $N$ of prime order $p$, for some $p|n$ Facts (1), (2) $G$ is supersolvable of order $n$ By the definition of supersolvability, $G$ has a nontrivial cyclic normal subgroup, say $C$. Let $p$ be a prime dividing the order of $C$. $C$ has a subgroup $N$ of order $p$, which by Fact (1) is characteristic in C. By Fact (2), $N$ is normal in $G$.
2 $G/N$ has subgroups of every order dividing $n/p$. inductive assumption By Fact (3), $G/N$ has order $n/p$, which is strictly smaller than $n$. Also, by Fact (4), it is supersolvable. Therefore, the inductive assumption applies, and it has subgroups of all orders dividing it.
3 If $p|d$, then $G$ has a subgroup of order $d$ Steps (1), (2) Since $d | n$ and $p | d$, we get that $d/p | n/p$. Therefore, $G/N$ has a subgroup of order $d/p$. By Fact (4), the inverse image of this in $G$ has order $d$.
4 If $p$ does not divide $d$, $G$ has a subgroup $K$ of order $pd$ containing $N$ as a cyclic normal subgroup of order $p$. Steps (1), (2) Since $d$ divides $n$ but $p$ does not divide $d$, $d | n/p$. Therefore, $G/N$ has a subgroup of order $d$. By Fact (4), the inverse image of this in $G$ has order $pd$. Call this group $K$.
5 If $p$ does not divide $d$, $G$ has a subgroup of order $d$. Step (4) Continuing from Step (4), $N$ is a normal Sylow subgroup of $K$ (since $p$ does not divide $d$). Hence, fact (5) tells us that $K$ contains a complement to $N$, which must therefore have order $d$.
6 In all cases, $G$ has a subgroup of order $d$ Steps (3), (5) The two cases were dealt with in steps (3) and (5).
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