Surjective morphism of automorphism group

Question

Let $A$ be a abelian group and $B$ a subgroup of $A$.
Is $Aut_B(A)\to Aut(A/B)$ surjective? Where $Aut_B(A)=\{\tau\in Aut(A)|\tau(B)=B\}$

Answer 1

Let $A=C_4\times C_2$, with the factors generated by $x$ and $y$, and let $B=\langle x^2\rangle\times\{1\}$. Then $A/B$ is isomorphic to the Klein $4$-group, so its automorphism group is isomorphic to $S_3$, which has order $6$.

On the other hand, the automorphism group of $A$ has order $\phi(4)\phi(2)2^{2} = 8$.. There is no surjection from any subgroup of a group of order $8$ to $S_3$, so the corresponding map cannot be a surjection.

If you want to see this directly (without refering to the order of $\mathrm{Aut}(A)$), note that there is an automorphism of $A/B$ that switches the image of $x$ and the image of $y$; but this cannot come from any automorphism of $A$, as the image of $y$ under any automorphism must be an element of order $2$ in $A$, and hence is contained in $B\times C_2$. So any automorphism of $A$ that sends $B$ to itself must induce a map on $A/B$ that fixes the image of $y$.