Finite quasisimple implies every endomorphism is trivial or an automorphism

Verbal statement
In a finite quasisimple group (i.e., a finite group that is also quasisimple), every endomorphism is either the trivial map, or an automorphism.

Quasisimple group
A group $$G$$ is termed quasisimple if $$G/Z(G)$$ is a simple group, and $$G$$ is a perfect group.

Note that $$G/Z(G)$$ is forced to be a simple non-Abelian group (if it were Abelian, $$G$$ would be solvable and hence not perfect). Thus, $$G/Z(G)$$ is centerless.

Facts used

 * 1) Proper and normal in quasisimple implies central: In a quasisimple group, any proper normal subgroup is contained in the center.
 * 2) Product formula
 * 3) Cocentral implies Abelian-quotient: If a subgroup, along with the center, generates the whole group, then it must contain the commutator subgroup.

Proof
Given: A finite quasisimple group $$G$$ with center $$Z(G)$$. An endomorphism of $$G$$ with kernel $$N \triangleleft G$$ and with image $$H \le G$$

To prove: $$N$$ is either trivial or the whole group $$G$$

Proof: Assume that $$N$$ is not the whole of $$G$$. Then, by fact (1), $$N \le Z(G)$$.

Clearly, $$Z(G)/N$$ is a proper subgroup of $$G/N$$ contained in the center of $$G/N$$. Moreover, the quotient $$(G/N)/(Z(G)/N)$$ is isomorphic to $$G/Z(G)$$, hence is simple.

Now the image of $$Z(G/N)$$ under the quotient map by $$Z(G)/N$$ is a proper normal subgroup of $$G/Z(G)$$, hence is trivial. The upshot: $$Z(G/N) = Z(G)/N$$.

Since $$H \cong G/N$$, we get:

$$\left |Z(H) \right | = \frac{\left|Z(G)\right|}{\left|N\right|}$$

Further, we have:

$$\left|H \cap Z(G) \right| \le \left| Z(H) \right|$$

Thus:

$$\left| H \cap Z(G) \right| \le \frac{\left|Z(G)\right|}{\left|N\right|} = \frac{\left|Z(G)\right|\left|H \right|}{\left| G \right|}$$

Rearranging, we get:

$$\left| H \cap Z(G) \right|\left| G \right| \le \left|Z(G)\right|\left|H \right|$$

Using the product formula (fact (2)), we get:

$$|G| \le |HZ(G)|$$

Since $$HZ(G) \le G$$, this forces:

$$HZ(G) = G$$

By fact (3), $$H$$ contains the commutator subgroup of $$G$$. But since $$G$$ is assumed to be perfect, $$H = G$$, forcing $$N$$ be be trivial, and completing the proof.