Ignoring elements of small order in the simple group of order $60$

Solution 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).