Finite quasisimple implies every endomorphism is trivial or an automorphism

Statement

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.

Definitions used

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.