The simple group of order $60$ can be generated by the permutations $(1,2)(3,4)$ and $(1,3,5)$, but all you need to do is square the first one and it becomes the identity. Can't we find a version of the simple group where the elements of small order can be ignored?

For a group $H$, define $Ω_n(H)$ to be the subgroup generated by elements of order less than $n$. For instance, if $n=3$ and $H=\operatorname{SL}(2,5)$ is the perfect group of order $120$, then $Ω_n(H)$ has order $2$, and $H/Ω_n(H)$ is the simple group of order $60$. If $n=4$ and $H=\operatorname{SL}(2,5)⋅3^4$ is the perfect group of order $(60)⋅(162)$ whose $3$-core is not complemented, then $Ω_n(H)$ has order $162$ and $H/Ω_n(H)$ is again the simple group of order $60$.

My first question is if there are smaller examples for $n=4$, since the jump $1$, $2$, $162$ seems a bit drastic for $n=2, 3, 4$.

Is there a group $H$ of order less than $(60)⋅(162)$ such that $H/Ω_4(H)$ is the simple group of order $60$?

Probably, for each positive integer $n$, there is a finite group $H$ such that $H/Ω_n(H)$ is the simple group of order $60$. I am interested in whether such $H$ can be chosen to be "small" somehow.

Is there a sequence of finite groups $H_n$ and a constant $C$ such that $H_n/Ω_n(H_n)$ is the simple group of order $60$ and such that $|H_n| ≤ C⋅n$?

I would also be fine with some references to where such a problem is discussed. It would be nice if there was some sort of analogue to the Schur multiplier describing the largest non-silly kernel, and a clear definition of what a silly kernel is (I think it is too much to ask for a non-silly kernel to be contained in the Frattini subgroup, and I think it might be unreasonable to ask for the maximum amongst minimal kernels).

In case it helps, here are some reduced cases that I know can be handled:

A simpler example: if instead of the simple group of order $60$, we concentrate on the simple group of order $2$, then we can choose $H_n$ to be the cyclic group of order $2^{1+\operatorname{lg}(n−1)}$ when $n≥2$, and the order of $H_n$ is bounded above and below by multiples of $n$. We can create much larger $H_n$ for $n≥3$ by taking the direct product of our small $H_n$ with an elementary abelian $2$-group of large order, but then $Ω_n(H_n×2^n) = Ω_n(H_n)×2^n$ has just become silly since the entire elementary abelian $2$-group part, $2^n$, is unrelated and uses a lot of extra generators, that is, it is not contained within the Frattini subgroup.

A moderate example: if instead of the simple group of order $60$, we take the non-abelian group of order $6$, then I can find a natural choice of $H_n$ with $|H_n| ≤ C⋅n$, but I am not sure if there are other reasonable choices. My choice of $H_n$ has $Φ(H_n)=1$, which suggests to me that Frattini extensions may not be the right idea.

  • 10
    $\begingroup$ Out of curiosity: why here and not MO? $\endgroup$ Aug 18, 2010 at 14:31
  • 1
    $\begingroup$ This post was an impressive display of math via HTML, but I felt like adding LaTeX instead :) Please double-check that I have not changed your meaning; there were a couple of places where I think I made minor corrections, but perhaps they are minor errors instead. $\endgroup$ Dec 1, 2011 at 7:43
  • $\begingroup$ You ask "Can't we find a version of the simple group where the elements of small order can be ignored?". However, elements of order $2$ (involutions) are very important in (non-abelian) simple groups, as it is a theorem that every group of odd order is soluble (and so not simple). Thus, in some respects, involutions hold the "key" to non-abelian simple groups. (I'm sure you know all this - I suppose I am wondering if you hope to find something interesting out of this, or is it just for fun?) $\endgroup$
    – user1729
    Dec 1, 2011 at 15:47

1 Answer 1


Regarding your first question about groups of order less than $162\cdot 60$ for which $H/\Omega_4(H)$ equals $A_5$, the simple group of order 60. Clearly, in any example the order of $H$ must be a multiple of 60; and also $H$ must be perfect. It is now a routine problem to write a GAP program, using the library of perfect groups, to find answers. Note that within the GAP system, each perfect group is identified by a pair $[n, i]$, where $n$ denotes the order, and the $i$ identifies which perfect group of that order is meant (in case there are multiple).

Using this, I verified that all perfect groups whose order is a multiple of 60 and less than $162\cdot 60$ satisfy $H/\Omega_4(H)=1$. So your example of order $9720=162\cdot 60$ is indeed the first where this quotient is non-trivial. And it is pretty special with that property, too; the next examples I found are of order $155520=2592\cdot60$ and $311040=5184\cdot60$. (But note that the database is incomplete for all orders $2^n\cdot60, n\geq 10$.)

The perfect group $[174960, 2]$ satisfies $H/\Omega_4(H)\cong A_6$. But for all other perfect groups up to the order $302400=5040\cdot 60$, the quotient is again trivial.

Then at order $311040=5184\cdot 60$ there are again a couple examples where the quotient is $A_5$.

And finally, the perfect group with id $[311040, 14]$ is the first (up to the gaps in the database!) group to satisfy $H/\Omega_5(H)\cong A_5$ (indeed, for all other groups before it, that quotient is trivial).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.