Finite solvable not implies subgroups of all orders dividing the group order

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finite solvable group) need not satisfy the second group property (i.e., group having subgroups of all orders dividing the group order)
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Statement

It is possible to have a finite solvable group $G$ of order $n$ and a divisor $d$ of $n$ such that $G$ has no subgroup of order $d$.

Related facts

Opposite facts

• Every finite solvable group is a subgroup of a finite group having subgroups of all orders dividing the group order
• Finite supersolvable implies subgroups of all orders dividing the group order
• Equivalence of definitions of finite nilpotent group, where one of the characterizations is that there exist normal subgroups of every order dividing the group order.

Proof

Further information: alternating group:A4, subgroup structure of alternating group:A4

Alternating group:A4 is the smallest counterexample. It has no subgroup of order six.

More generally, for any prime power $q=p^{r}$ with $r>1$, the general affine group $GA(1,q)$ obtained as the semidirect production of the additive and multiplicative groups of the field with $q$ elements is supersolvable but does not have subgroups of every order dividing the group order. In fact, we can show that it does not have any subgroup of order $p^{s}(q-1)$ for $0<s<r$. This is because any such subgroup must contain a nonzero element of the additive subgroup, and it also contains representatives of all the cosets for the multiplicative group, which in turn forces it to contain the entire additive group.