Example of finite groups with no subgroup for only one d such that $d | d_1$ and G has proper subgroup of order $d_1$.

Question

$S_5$ has no subgroup of order 15, 30, 40 and has a subgroup of order 60. So here we have two divisor that divide a order of a subgroup and 40 does not divide any.

Answer 1

I'm still not a hundred percent sure whether I fully understand the requirements – I believe you are looking for a finite group $G$ such that the following two conditions hold:

$$\text{For all maximal proper divisors $k$ of $|G|$, there exists a subgroup $K$ of $G$ with }|K|=k\tag{1}$$ $$\text{There exists a unique divisor $h$ of $|G|$ such that $G$ has no subgroup of order $h$.}\tag{2}$$

The search for such groups $G$ is not trivial. One obvious almost-example is the alternating group $A_4$ which has no subgroup of order $h=6$ and thus satisfies condition (2), but since $6$ is a maximal proper divisor of $|A_4|=12$, condition (1) is violated.

However, one can generalize this a bit by going from a semi-direct product of a two-dimensional vector space over $GF(2)$ (which is one way to look at $A_4$) to a three-dimensional space:

Let $V$ be the elementary abelian group of order $8$, and let $\alpha$ be an automorphism of $V$ of order $7$. Let $H$ be the semi-direct product of $V$ and $\langle\alpha\rangle$. Then $H$ has order $56$ and contains subgroups of all possible orders except $14$ and $28$.

Now, let $G$ be the direct product of $H$ and a group $T$ of order $2$. Then, $G$ does contain subgroups of orders $14$ (namely $\langle\alpha\rangle\times T$) and $56$ (namely $H\times 1$), but no subgroup of order $28$, and it is easy to check that $28$ is in fact the only divisor of $|G|=112$ for which $G$ has no subgroup of that order.

So, $G$ would be an example of a group you are looking for.