Is there hypothesis such that the reciprocal of Lagrange theorem is true (except sylow theorem) If have in my course the following theorem

$\textbf{Lagrange}:$ Let $G$ a finite group and $H$ a subgroup of $G$. Then the order of $H$ divide the order of $G$.

I know that the reciprocal is wrong, i.e. if $k$ divide the order of $G$, there is not necessarily a subgroup of $G$ of order $k$. Nevertheless, for $\mathbb Z/n\mathbb Z$ it hold, i.e. if $k\mid n$ then there is a subgroup of order $k$ (for example $\mathbb Z/k\mathbb Z$ is a subgroup of $\mathbb Z/n\mathbb Z$). Is there such a result for more general group ? (i.e. not only $\mathbb Z/n\mathbb Z$) ? Like maybe commutative group is enough to get the result ?   
 A: It is true for finite supersolvable groups. That is, groups with a normal series  with cyclic factor groups. This includes all nilpotent groups.The proof is an easy induction on the group order. Nontrivial finite supersolvable groups $G$ have a normal subgroup $N$ of prime order $p$, and you just apply induction to $G/N$ and use the Schur-Zassemhaus Theorem for subgroups of order not divisible by $p$.
Conversely, a group with subgroups of every possible order is solvable. That follows from a more general result of P. Hall, that groups with Hall $\pi$-subgroups for all sets of primes $\pi$ are solvable. But it is not true for all solvable groups (e.g. $A_4$). On the other hand there exist non-supersolvable groups  for which it is true, like $A_4 \times C_2$, and I don't think there is a precise classification of groups with thisproperty.
A: Be careful, $\mathbb Z/k\mathbb Z$ is not a subgroup of $\mathbb Z/n\mathbb Z$, elements in $\mathbb Z/k\mathbb Z$ are sets of the form $z+k\Bbb Z$, $z\in \Bbb Z$, while elements in $\mathbb Z/n\mathbb Z$ are sets of the form $z+n\Bbb Z$, $z\in \Bbb Z$.
I think that what you were thinking is the following

$k\mathbb Z/n\mathbb Z$ is a subgroup of $\mathbb Z/n\mathbb Z$,

which is correct. The generalization of what you're thinking is the following

Let $G$ be a finite cyclic group of order $n$, then for each $k$ dividing $n$, there is a subgroup $H$ of order $k$. More specifically, if $G=\langle g \rangle$, such a subgroup $H$ is equal to $\langle g^{\frac{n}{k}}\rangle$

If we apply that result to the case $G=\Bbb Z/n\Bbb Z=\langle 1+n\Bbb Z\rangle$, we see that the subgroup of order $k$ is $H=\langle k\cdot 1+n\Bbb Z\rangle=k\Bbb Z/n\Bbb Z$
