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Let $G$ be a group, with $N$ characteristic in $G$. As $N$ is characteristic, every automorphism of $G$ induces an automorphism of $G/N$. Thus, $\operatorname{Aut}(G)\rightarrow \operatorname{Aut}(G/N)$. I was therefore wondering,

Under what conditions is the induced homomorphism $\operatorname{Aut}(G)\rightarrow \operatorname{Aut}(G/N)$

  • a monomorphism?

  • an epimorphism?

  • an isomorphism?

I believe it should work for (semi-?)direct products $N\times H$ where $\operatorname{Aut}(N)$ is trivial and $N\not\cong H$ (for example, $C_2\times C_3$, $N=C_2$). But I can't prove even that!

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    @user1729 The 3 questions you asked for general pairs (G,N) is an extremely hard question to address, such behaviour is for the most part only known in nice classes of groups. In particular, if$G$is d-generated and finite (d the size of a smallest generating set for G), and S=S_d(G) is the collection of all ordered generating d-tuples, the diagonal action of G on S is a free action, but not transitive in general. If it IS transitive, like for G=C_n, d=1 or G=Quaternions d=2, and d-generating tuple for$G/N$lifts to one for G! This is an old Thm of Gasch\"utz's. Then |Aut(G)|=|S| and .... well,2012-06-21

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For direct product it seems easy.

Let $G=N\times K$ and assume that both $N$ and $K$ are characteristic in $G$. It is easy to show $Aut(G) \cong Aut(N)\times Aut(K)$ since both $N$ and $K$ are characteristic in G. Since $G/N \cong K$ then $Aut(G/N) \cong Aut(K)$. Thus there is a natural epimorphism $\phi:Aut(G) \to Aut(G/N)$ with $ker(\phi) \cong Aut(K)$.

Now you can ask when are they both characteristic in $G$? Actually, one simple condition provide this: Let $N$ and $K$ be finite groups with relatively prime orders, and set $G=N\times K$. Then both $N$ and $K$ are characteristic in $G$.

And you offer an example $G=C_2\times C_3$ and $N=C_2$ then set $K=C_3$ since order of N and K are relatively prime, the result is immediate from above construction.