I am trying to find a streamlined proof that $A_5$ is the smallest perfect group, with respect to order ( a group is perfect if it has no abelian quotients except $\{e\}$). I know it follows directly from the classification of simple groups, as it is the smallest non-abelian simple group. Is there a way to obtain this result without too much work without using this tool?
Thanks in advance, this question is just for fun though, I really want to see what techinques we can come up with.
A perfect group that is not simple has a nontrivial perfect quotient. As a result, finding the smallest nontrivial perfect group is equivalent to finding the smallest non-abelian simple group.
As P Vanchinathan mentions, we can use Burnside to eliminate all cases except order $30$. A group of order $30$ with no abelian factor is isomorphic to a dihedral group, but every dihedral group has a subgroup of index 2. But Burnside is a quite powerful result, with no elementary proof that I know of.
The usual elementary arguments proceed by case analysis, using the Sylow theorems. Here is an outline of a typical argument, for example. Some cases are easy, while others require some work. For example, I don't know of an easier way than the Sylow theorems and counting arguments to show that no non-abelian group of order $56$ is simple.