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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 $G\cong H$?

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    $\begingroup$ You might consider checking your question against the data here: madore.org/~david/math/simplegroups.html $\endgroup$
    – Neil Hoffman
    Commented Apr 24, 2014 at 9:53
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    $\begingroup$ A computer calculation shows that the simple groups of orders up to $2000$ are all characterized by their subgroup orders. It seems to be a reasonable conjecture that this might be true in general, but it is hard to think how you might go about proving it. $\endgroup$
    – Derek Holt
    Commented Apr 24, 2014 at 11:02
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    $\begingroup$ The property also holds for many non-simple groups, e.g., groups of order $pq$ with $p<q$ primes and $q-1$ not a multiple of $p$. $\endgroup$
    – Gerry Myerson
    Commented Apr 24, 2014 at 12:39
  • $\begingroup$ How do you handle it when there are many suggroups of the same order. If you discount multiplicity the two groups of order 4 will have the same 'horoscope'. $\endgroup$
    – P Vanchinathan
    Commented Apr 24, 2014 at 13:23
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    $\begingroup$ Generalizing the comment of @GerryMyerson this holds for any group whose order is a cyclic number (and these can be recognized from the set of subgroup orders by these all being square free and the product of the set of their prime divisors being a cyclic number). $\endgroup$
    – Tobias Kildetoft
    Commented Apr 25, 2014 at 7:19

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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), 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 $\langle x,y \mid x^8=y^2=1, x^y=x^5 \rangle$.

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      $\begingroup$ Out of curiosity, wouldn't you want to weigh each group $G$ by $1/Aut(G)$ in the count of probabilities? $\endgroup$
      – Lev Borisov
      Commented Apr 25, 2014 at 0:28
    • $\begingroup$ What means $1/Aut(G)$? $\endgroup$
      – Marius Tarnauceanu
      Commented Apr 25, 2014 at 0:59
    • $\begingroup$ Lev Borisov: Why would you want to do this? My understanding of an 'almost all' statement about finite groups is that the probability of a group in the class of groups of order at most $n$ (chosen uniformly amongst the isomorphism types) satisfying the statement tends to $1$ as $n$ tends to infinity. $\endgroup$
      – Colin Reid
      Commented Apr 25, 2014 at 2:49
    • $\begingroup$ @ColinReid, of course one can weight isomorphism classes however one wants, but I think that @‍LevBorisov's suggestion was that perhaps groups with lots of automorphisms should be weighted less than more rigid groups. This is in line with groupoid cardinality, but whether or not it is more natural is probably not mathematically well defined. $\endgroup$
      – LSpice
      Commented Jul 28, 2021 at 21:43

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