Finite quasisimple implies every endomorphism is trivial or an automorphism

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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.


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|


\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.