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$S_5$ 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.

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    Can you clarify exactly what you're asking for? You want a finite group $G$ such that there is exactly one pair $(d,d_1)$ such that $G$ has a proper subgroup of order $d_1$, has no subgroup of order $d$, and $d\mid d_1$? Or is only $d$ supposed to be unique (and there must exist at least one $d_1$ that works with it), rather than the pair $(d,d_1)$? Or something else? – 2017-02-20
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    only d supposed to be unique (and there must exist at least one d1 that works with it). – 2017-02-20

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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.

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    Thanks you very much for your reply. In GAP, I found that SmallGroup(112,41); has this property. I checked for the group of order $3^4\times 7$ and $2^4\times 5$ or $2^4 \times 11$. They are not satisfying this property. Can we say in general for what n we will going to get this type of property. – 2017-02-22
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    @SSSM: This is an interesting question. If you look at the lattice of divisors of $H$ (in the example above), you see that there is a "chain" $7 - 14 - 28 - 56$ of length $4$ in on "direction" (represented by the common prime quotient $2$) such that the first and last members of that chain have subgroups with these orders but the two middle elements don't. In a case like this, taking the direct product with a cyclic group of the order of that quotient will "cover" one of the missing divisors but leave the second missing; that way you'll reduce the number of "holes". – 2017-02-23
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    (comment continued:) So, looking for constellations like these will yield further examples. Of course, there may be other mechanisms to construct groups like the ones you are looking for, so a "chain" like the one above is most likely not a necessary condition to arrive at another example. – 2017-02-23