Question

Let G=A⋊B, where A and B are abelian, and of coprime order. It seems, from my computations (and correct me if I'm wrong), that Z¹(C_p,C_q) is trivial, for p and q different primes. Meaning that the automorphisms of G, if A=C_q, and B=C_p, that preserve A, and preserve the cosets G/A, are all trivial. How far can we extend this? Would it be true in general that Z¹(A,B) is trivial, with the above assumptions (that A and B are abelian and of coprime order)? If not, under what assumptions is it trivial? And when can we say about it if it's not trivial?

Answer 1

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.

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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 $-j$.

Answer 2

You might be considering a special case of the Schur-Zassenhaus theorem. If A is a normal Hall-subgroup of G, then A has a complement B, and all complements B are conjugate under the action of G. This is more properly H^1(B,A) rather than Z^1(B,A).

For Z^1(B,A) to be trivial, B^1(B,A) must be trivial, but B^1(B,A) is basically [B,A], which could be nonzero. For instance if G is non-abelian of order 6, then A is cyclic and normal of order 3, B is cyclic of order 2, and B acts as inversion on A. Then for B^1(B,A) should be isomorphic to A, that is have order 3. Basically, A has 3 complements in G, so B^1(B,A) should have three elements, but all are conjugate, so H^1(B,A)=0.

I suppose there are lots of definitions of C^1(B,A), so maybe you've chosen one where B^1(B,A)=0, but I think the standard choice when looking at semi-direct products ("Crossed homomorphisms"), will not have B^1(B,A) trivial.