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!
Best Answer
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.