I have a general question about automorphism groups. Sorry in advance if I'm talking about well known facts, but I didn't find much in the literature.

Let $G$ be a group and let $N$ be a characteristic subgroup of $G$ (that is, $\varphi(N)=N$ for all $\varphi \in \text{Aut}(G)$). Then we have a natural map $f: \text{Aut}(G) \to \text{Aut}(G/N)$. What can be said about this map in general? What about its kernel? When is it surjective?

My intuition leads me to think that if $N$ is reasonably "small" (for example, contained in the derived subgroup, in the center and in the Frattini subgroup) then $f$ is an isomorphism. Is this really reasonable?

I'm thinking for example at the (apparently) easiest case: take $N$ to be the unique normal subgroup of $G$ of order $2$ (if such $N$ exists $G$ is usually called "binary" - right? - cf. http://cameroncounts.wordpress.com/2011/06/22/groups-with-unique-involution/).

Also, reading the paper "H.K. Iyer, "On solving the equation Aut(X)=G", Rocky Mountain J. Math. 9 (1979), no. 4, 653--670" (pages 7-8) I see that $f$ is an isomorphism if $G$ is the universal cover of $G/N$ and $G/N$ is simple.

Thank you for any contribution.

  • $\begingroup$ There is a very similar question on SE, math.stackexchange.com/questions/64492/… $\endgroup$
    – ADL
    Jul 28, 2012 at 19:57
  • 1
    $\begingroup$ Exercise 11.4 in Derek Robinson's book "A Course in the Theory of Groups" (2nd edition) determines the kernel $K$ of the map $Aut(G)\to Aut(N)\times Aut(G/N)$ in case that $C:=C_G(N)$ equals the center $Z(N)$ of $N$: This kernel $K$ is an abelian group isomorphic to the group of derivations $\mathop{Der}(G/N, C)$ and $K/K\cap\mathop{Inn}(G)$ is isomorphic to $H_1(G/N,C)$. $\endgroup$
    – j.p.
    Jul 29, 2012 at 17:32

1 Answer 1


It's not reasonable: take $G$ to be a finite non-abelian group of order $p^3$ for prime $p$, and $N$ its derived subgroup (which is also its center and its Frattini subgroup, and has order $p$); then any nontrivial inner automorphism of $G$ in a nontrivial element in the kernel of $f$. There are some other examples of order $p^5$ (with derived subgroup of order $p$) for which $f$ is not surjective.


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