Common Computations in Group Cohomology

The automorphisms of this extension are basically the same as group cohomology $H^1(B; A)$, so I will focus on that first.

So we want to show that this must be zero given that B and A are finite abelian and of coprime order. This is the same as the twisted cohomology of the classifying space BB. Now we can look at the bundle $p:EB to BB$. This is a covering space with fiber the discrete space B.

Now because the fiber is discrete we have a wrong way transfer map in twisted cohomology:

$$p_{!}: H^1(EB; A) to H^1(BB; A)$$

where the first group is twisted cohomology in the pulled-back local coefficient system. As with all transfers we have that

$$p_{!} circ p^*: H^1(B; A) to H^1(B; A)$$

is multiplication by the order of the fiber, i.e. $|B|$. Since the orders of A and B are coprime this is an isomorphism. But since $EG simeq pt$ is contractible, this map factors through the zero group and hence $H^1(B; A) = 0$.

Any proof that the cohomology of a group is torsion for the order of the group (there are more concrete ones then the above) will yield the same result that $H^1(B; A) = 0$. There are many ways to prove this (as the comments point out), the above is just my favorite.

---

So what is the difference between the $H^1(B;A)$ and the isomorphisms of the extension G? Well as you pointed out the isomorphisms of G (which restrict to the identity on A and the quotient B) are the same as the bar resolution coycycles Z^1(B;A). So we have an exact sequence,

$$A= C^0(B;A) to Z^1(B; A) to H^1(B;A) to 0$$

but as we saw, the term $H^1(B;A) = 0$. So we must compute the boundaries. You have one such potential homomorphism for each element of $A$, although different elements might give rise to the same automorphism of $G$. They are of the form:

$$b mapsto a - b cdot a$$

where $a in A$ is fixed. If the action of B on A is trivial, then these vanish, but in general they can be non-zero. My favorite example is the quaterion group which we view as

$$mathbb{Z}/4 to Q_8 to mathbb{Z}/2$$

with the $mathbb{Z}/4$ the group $( 1, i, -1, -i )$. An element $x in A = mathbb{Z}/4$ induces the homomorphism

$$y mapsto x - y cdot x$$

which sends the non-trivial element of $mathbb{Z}/2$ to $2x$. In particular it is non-trivial for a generator of A. This corresponds to the isomoprhism of $Q_8$ which sends $i$ to $i$ and $j$ to $-j$.