I have a conjecture that a certain criterion is enough for two groups to be isomorphic. I tested it on all pairs of groups up to size 12, and it worked like a charm. I know, however, that groups are strange and that it is very likely that my conjecture will break for larger groups. I have made myself a python library that handles making groups and seeing they have certain properties/are isomorphic, but I do not know what pairs of groups to test my conjecture on.



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> What groups form a "representative sample" of all groups? What paris of groups would it be absolutely crucial for me to test my hypothesis on?
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I don't see the need for any answer to be complete, but any ideas for pairs that would be interesting to test on would be appreciated.


EDIT:


The conjecture is that if two groups have the same count of elements of every order, then they must be isomorphic. I proved this for abelian groups, and am now wondering if it is true in general.



What groups should I test my conjecture on?
Your conjecture is false. Probably an explicit counterexample is easy to write down but here's an existence proof that counterexamples are plentiful: asymptotically it's [known](https://en.wikipedia.org/wiki/Higman%E2%80%93Sims_asymptotic_formula) (Higman-Sims) that there are $p^{ \frac{2}{27} n^3 + O(n^{8/3})}$ groups of order $p^n$, for $p$ a prime. The elements of such a group have one of $n+1$ possible orders $1, p, \dots p^n$, and so the number of possible different order profiles of such a group is at most the number of compositions of $p^n$ into at most $n+1$ parts, which is ${p^n + n \choose n}$ which only grows at best like $O(p^{n^2})$ (edit: and see the comments for more on this). So asymptotically there are many more groups of order $p^n$ than there are possible order profiles.


**Edit:** [This MO answer](https://mathoverflow.net/a/39850/290) contains the explicit counterexample of the Heisenberg group $H\_3(\mathbb{F}\_3)$ and $C\_3^3$ (order $27$, so $p = 3, n = 3$), which are not isomorphic but have the same order profile (all non-identity elements have order $3$). [This MO answer](https://mathoverflow.net/a/223820/290) to the same question says the smallest counterexamples have order $16$.


**Edit #2:** It is maybe worth saying that the fast growth of the number of groups of order $p^n$ kicks in already for pretty small values of $n$. For $n = 10$ there are about $49$ million groups of order $2^{10} = 1024$ and these account for over $99\%$ of the groups of order $\le 2000$. See, for example, [this blog post](https://golem.ph.utexas.edu/category/2012/11/almost_all_of_the_first_50_bil.html). Most of these groups are $2$-step nilpotent so have elements of orders $1, 2, 4$, and so as YCor says in the comments these groups have at most $(2^{10} - 1) + 1 = 1024$ order profiles, so by pigeonhole we know at least some collection of $\approx 49000$ of these groups have the same order profile. Conjecturally "almost all" finite groups are $2$-step nilpotent $2$-groups so this is in some sense "typical."


