Ignoring elements of small order in the simple group of order $60$ 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.
 A: 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).
