# Finite quasisimple implies every endomorphism is trivial or an automorphism

## Contents

## 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 is termed **quasisimple** if is a simple group, and is a perfect group.

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.

## Facts used

- Proper and normal in quasisimple implies central: In a quasisimple group, any proper normal subgroup is contained in the center.
- Product formula
- 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 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:

Thus:

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.