TITLE: Common Computations in Group Cohomology
QUESTION [7 upvotes]: 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(Cp,Cq) is trivial, for p and q different primes. Meaning that the automorphisms of G, if A=Cq, and B=Cp, 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?

REPLY [3 votes]: 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.