This is related to If $G$ and $H$ are nonisomorphic group with same order then can we say that $Aut(G)$ is not isomorphic to $Aut(H)$? and Can non-isomorphic abelian groups have isomorphic endomorphism rings? but more general than both. The answer given in the first link says that two finite non-isomorphic groups can have isomorphic automorphism groups. The second link (apparently) gives an example of two non-isomorphic infinite groups with isomorphic endomorphism rings. But if $G$ and $H$ are two non-isomorphic finite groups of the same order, is it possible for their endomorphism monoids to be isomorphic?


1 Answer 1


Not necessarily.

I cannot think of a simple example right now, but there are plenty of examples of finite $p$-groups $G$ of nilpotence class $2$ such that

  1. all automorphisms are central, and
  2. an endomorphism that is not an automorphism maps $G$ into $Z(G)$.

Now if you take two non-isomorphic such groups $G_{1}, G_{2}$ in which the $G_{i}/Z(G_{i})$ and the $Z(G_{i})$ are elementary abelian and isomorphic, then their endomorphism monoid are isomorphic.

As a reference, see this paper of mine (should be freely accessible, in case please advise), particularly Section 2 and Theorem 4.3.


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