Is there any finite groups with no subgroup for one and only one divisor of order of G of index greater than 2. $A_4$ has no subgroup of order 6(only one) but index is 2.
Also $A_5$ has no subgroup of index 2,3 and 4.
I am unable to find an example of a finite group with no subgroup of unique index other than 2.
 A: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Let $p, q$ be primes, with $q \nmid p -1$, but $q \mid p^{2} - 1$.
Let $F$ be the field with $p^{2}$ elements. Since $q \mid p^{2} - 1$, in the multiplicative group $F^{*}$ of order $p^{2} -1 $ there is an element $g$ of order $q$.
Consider the semidirect product of the additive group of $F$ by $\Span{g}$,with $g$ acting on $F$ by multiplication.
This has order $p^{2} q$, it has subgroups of index $q$ and $p^{2}$ (a Sylow $p$-subgroup resp. a Sylow $q$-subgroup), but not of index $p$. The reason is that such a group $H$ would have order $p q$. Since $p > q$, a subgroup $N$ of order $p$ of $H$ would be normal in $H$. Since $q \nmid p - 1$, $H$ would turn out to be abelian. This means there is an element of order $q$ centralizing an element of order $p$, but since $g$ acts by multiplication, this is not the case.
The case of $A_{4}$ is the special case $p = 2$, $q = 3$. The next case is $p = 5$, $q = 3$.
A: The group $$(C5\times C5):C9$$ has no subgroup of order $45$ (Index $5$) , but at least one subgroup of any other order dividing $225$.
