Perfect not implies semisimple

Statement
It is possible for a group (in fact, a finite group) to be a perfect group (i.e., it equals its own derived subgroup) but not a semisimple group (i.e., it is not expressible as a central product of quasisimple groups, or equivalently, its layer is a proper subgroup of it).

Proof
The following is one recipe for constructing counterexamples. Let $$n \ge 5$$. Take the wreath product with base cyclic group:Z2 and acting group the alternating group $$A_n$$. This is a group of order $$2^n(n!)/2$$. The center is cyclic of order two. Quotient out by it and get the inner automorphism group. This is a perfect group of order $$2^{n-2}n!$$ and is not semisimple.

The smallest example is the inner automorphism group of wreath product of Z2 and A5, which is a group of order 960. There is another similar example of order 960. 960 is the smallest order for any example.