Is every finite group the quotient of a finite group by the subgroup generated by elements of small order?

For a positive integer n, define Ω(H) to be the subgroup of H generated by those elements of order less than n. Is it true that for every finite group G there is a finite group H and normal subgroup Ω(H) ≤ N ⊴ H with H/N ≅ G?

So it’s pretty easy to do for G = Sym(3), one just takes O^{2′}(Hol(3^{k})) for large enough k.

I guess it is a little surprising that G = Z/2Z is easy to do: after all H = Z/6Z does not work for n = 4, since Ω(H) = H. However, H = Z/4Z works just fine, since Ω(H) = 2H, and H/2H = G.

So one cannot just arbitrarily lift elements to higher orders. Some sort of prime power lifting is needed, but what do you do with elements whose order has more than one prime divisor?

Posted at math.se#1919:

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=SL(2,5) is the perfect group of order 120, then Ω_{3}(SL(2,5)) has order 2, and H/Ω_{n}(H) is the simple group of order 60. If n=4 and H=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+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.