Given a finite group $G$, we denote by $\pi\_s(G)$ the set of orders of its subgroups. Which finite groups $G$ can be characterized by the set $\pi\_s(G)$, i.e. $\pi\_s(H)=\pi\_s(G)$ implies $H\cong G$? Elementary examples of such groups are cyclic groups of prime order, $A\_4$, ... and so on. Is this property true for finite simple groups? (similarly with the recognition of finite simple groups by their spectrum). Any suggestions or references are welcome.


Additional question: Let $G$ and $H$ be two finite groups. Assume that $\pi\_s(G)=\pi\_s(H)$ and $card\{K\leq G \mid |K|=d\}=card\{K\leq H \mid |K|=d\}, \forall\, d\in\pi\_s(G)$. Is it true that $G\cong H$?



Which finite groups can be characterized by their subgroup orders?
This property is rare in groups of small order. If I have calculated correctly then, of the $1237$ group of order at most $120$, $56$ have this property. But for $44$ of these, there is a unique group of that order, so there are really only $12$ interesting examples, which include $A\_4$, $A\_5$, ${\rm SL}(2,5)$, but not $S\_5$, which has the same subgroup order type as $A\_5 \times C\_2$.


The principal reason that the property is rare is that a large majority of groups appear to satisfy the converse of Lagrange's Theorem: they have subgroups of all orders dividing the group order.


It would be very nice if there were statements of that kind that could be proved, but at present the available techniques do not seem to be adequate for this purpose.


Here is a sequence of reasonable conjectures of increasing strength:

- Almost all finite groups satisfy the converse of Lagrange's Theorem.
- Almost all finite groups are supersolvable.
- Almost all finite groups are nilpotent.
- Almost all finite groups are $2$-groups.
- Almost all finite groups are $2$-groups of nilpotency class $2$ - you can add extra conditions to this.



There are asymptotic estimates on number of groups with various properties
(see for example [Number of isomorphism types of finite groups](https://mathoverflow.net/questions/151491)), but they are not precise enough to prove any of these "almost all" statements.


For your final question, the answer is of course no. It is never possible to distinguish between isomorphism types of groups using numerical data of this type. So you are really just asking someone to do a search for the smallest counterexample, which is fortunately not too hard in this case: there are two pairs of groups of order $16$. One of the two pairs is $C\_4 \times C\_4$ and $\langle x,y \mid x^4=y^4=1, x^y=x^{-1} \rangle$, and the other pair is $C\_2 \times C\_8$ and $\langle x,y \mid x^8=y^2=1, x^y=x^5 \rangle$.


