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.