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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.

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$A_5$ is a simple group of order $60$. It therefore cannot have a subgroup of order $30$, since subgroups of index $2$ are always normal.

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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.

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    $\begingroup$ I like that your answer contained the smallest possible example. +1 $\endgroup$ – user729424 Jan 31 '20 at 17:04

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