Finite quasisimple implies every endomorphism is trivial or an automorphism
Note that is forced to be a simple non-Abelian group (if it were Abelian, would be solvable and hence not perfect). Thus, is centerless.
Given: A finite quasisimple group with center . An endomorphism of with kernel and with image
To prove: is either trivial or the whole group
Proof: Assume that is not the whole of . Then, by fact (1), .
Clearly, is a proper subgroup of contained in the center of . Moreover, the quotient is isomorphic to , hence is simple.
Now the image of under the quotient map by is a proper normal subgroup of , hence is trivial. The upshot: .
Since , we get:
Further, we have:
Rearranging, we get:
Using the product formula (fact (2)), we get:
Since , this forces:
By fact (3), contains the commutator subgroup of . But since is assumed to be perfect, , forcing be be trivial, and completing the proof.