# A group $G$ with $n$ elements has the property $(L)$ if for any divisor $d$ of $n$ there is at least a subgroup $H$ of $G$ with $d$ elements

A group with $$n$$ elements has the property $$(L)$$ if for any divisor $$d$$ of $$n$$ there is at least a subgroup $$H$$ of $$G$$ with $$d$$ elements. Prove that there are finite groups which have the property $$(L)$$ and that there are also some which do not have it.
Now, it is easy to see that $$K_4$$(the Klein four group) and $$Z_n$$($$n\in \mathbb{N}$$) have the property $$(L)$$. I couldn't think of a group which doesn't.

$$A_5$$ is a simple group of order $$60$$. It therefore cannot have a subgroup of order $$30$$, since subgroups of index $$2$$ are always normal.
The group $$A_4$$ has no subgroup of order $$6$$. Assume to the contrary that $$H\subset A_4$$ is a subgroup of order $$6$$. As such, we either have $$H\cong\mathbb{Z}/6\mathbb{Z}$$ or $$H\cong S_3$$. The former can't happen, because $$S_4$$ has no elements of order $$6$$. If $$H\cong S_3$$, then it is generated by an element of order $$2$$ and an element of order $$3$$. In $$A_4$$, an element of order $$2$$ is a double transposition and an element of order $$3$$ is a $$3$$-cycle. Together, these act transitively on the set $$\{1,2,3,4\}$$, contradicting the orbit-stabilizer theorem ($$4\nmid6$$). Therefore, no such subgroup $$H$$ exists. This is, in fact, the unique example of a group with minimal order not having the required property.