an Equivalent condition for nilpotency of a finite group like of finite solvable group

Question

We know that these two conditions are equivalent for finite groups :

i) $G$ is solvable.

ii) If $\left|G\right|=mn\;$ and $(m,n)=1$ then G has a subgroup of order $m$.

Now are the following two conditions equivalent for finite groups? We know that i)$ \implies$ ii) holds.

i) $G$ is nilpotent.

ii) If $m$ divides $|G|$ then $G$ has a subgroup of order $m$

Answer 1

I find a counterexample. $S_3$ has subgroups of order 1,2,3 and 6 but isn't direct product of it's sylow p-subgroups and hence isn't nilpotent.

Answer 2

The collection of all finite groups $G$ for which there exists a subgroup of order $n$ for all $n\mid |G|$ are called CLT-groups (converse to Lagrange's theorem). Such groups lie strictly between the collection of supersoluble and soluble groups.

The correct equivalent condition for $G$ being nilpotent is that $G$ possesses a normal subgroup of each order dividing $|G|$. To prove this, $p$-groupshave normal subgroup of each order, and now use the fact that a nilpotent group is a product of its Sylow subgroups. This proves one direction. The other is that if a group has a normal subgroup of each order, then its Sylow subgroups are normal, and thus $G$ is nilpotent.