A subgroup of index $n$ Let $G$ be a group such that $G$ has not any subgroup of index $2,3,\ldots, n-1$. If $G$ has a subgroup of index $n$ such $H$, then whether $H$ is normal in $G$? 
 A: $A_4$ is a subgroup of $A_5$ with minimal index $5$ but it is not normal in $A_5$.
$\mathbb{Z}_2$ is a subgroup of $\mathbb{Z}_4$ with minimal index $2$ but it is normal in $\mathbb{Z}_4$.
Does this answer your question?
A: This is not really an answer, since the $A_4<A_5$ example by @muzzlator already shows that a subgroup meeting the conditions need not be normal. I will however show that this example is minimal, in the sense that the result holds true when $n\leq3$, and also vacuously when $n=4$.
It is well known that index $2$ subgroups are always normal (and index $1$ too of course). The most instructive proof of this gives something for higher index groups as well. Let $G$ act on the set $G/H$ of cosets by left-multiplication, an action that is clearly transitive (the coset $gH$ is reached from $H$ using $g$). This gives a group morphism $G\to\mathcal S(G/H)$ from $G$ to the permutations of the set $G/H$, whose kernel $K$ is a normal subgroup of contained in $H$, and the index $[G:K]$ divides the order $n!$ of $\mathcal S(G/H)$ where $n=[G:H]$. When $n\leq2$ this suffices to prove that $K=H$, so $H$ is normal in $G$. If $n=3$ then either $K=H$ and $H$ therefore normal, or $[G:K]=6$ and the morphism $G\to\mathcal S(G/H)\cong S_3$ therefore surjective. But in the latter case the inverse image of $A_3<S_3$ is an index $2$ subgoup of $G$; if such subgroups do not exist then we must be in the former case.
Finally I'll show that $n=4$ cannot occur, i.e., there cannot be an index $4$ subgroup without the existence of either an index $2$ or an index $3$ subgroup (and the group $A_4$ shows that only having an index $3$ subgroup here is a possibility). As above the existence of $H$ with $[G:H]=4$ gives rise to a group morphism $G\to S_4$. If the image of this morphism is not contained in$~A_4$ then the inverse image of$~A_4$ gives an index $2$ subgroup of$~G$, and we have assumed that this does not happen, so we in fact have a morphism $G\to A_4$. Similarly if the image is not contained in the Klein $4$ subgroup $K_4$ of $A_4$, then taking the inverse image of $K_4$ gives an index $3$ subgroup; therefore we get a morphism $G\to K_4$. But the action being transitive on $4$ elements, this morphism must be surjective, but then the inverse image of any one of the index $2$ subgroups of $K_4$ gives an index $2$ subgroup of $G$, contradiction.
A: I'm not sure whether this is worth writing as an answer, rather than a comment, but for every non-Abelian finite simple group $G,$ there is a positive  integer $n>1$ such that $G$ has no proper subgroup of index less than $n,$ but $G$ contains a subgroup $H$ of index $n$ which is not normal in $G.$  Simply take a proper subgroup $H$ of $G$ with $[G:H]$ minimal and choose $n$ as $[G:H]$. Then $H$ is not trivial ( for $G$ has non-trivial cyclic proper subgroups) and $H \neq G,$ since $n > 1.$ Hence $H$ is certainly not normal in the simple group $G,$ and $G$ has no proepr subgroup of index less than $n$ by construction.
 For every integer $n >4,$ the alternating group $A_{n}$ is well-known to be non-Abelian simple, and it has no proper subgroup of index less than $n,$ but does have a subgroup of index $n,$ so for every integer $n$ greater than $4,$ there is a finite group $G$ which has no-proper subgroup of index less than $n,$ but which has a subgroup $H$ of index $n$ which is not normal. ( The reason that the alternating group $A_{n}$ has no proper subgroup index less than $n$ is that if there were such a subgroup, say of index $k,$ then there would be a group homomorphism from $G$ to $S_{k}$ whose kernel is not all of $A_{n}.$ Then the kernel must be trivial as $A_{n}$ is simple. But the image can't be contained in $A_{k}$ as $k <n,$  and the image is isomorphic to $A_{n}.$ If the image is not contained in $A_{k}$ however, then the image has a normal subgroup if index $2,$ contrary to the fact that the image is isomorphic to $A_{n}).$ 
A: The answer is affirmative for finite nilpotent groups since $H$ is maximal
http://en.wikipedia.org/wiki/Nilpotent_group.
