Bourbaki's definition of semidirect product This recess I'm off to learn about group extensions and the cohomological methods to characterize those extensions, but I'm a bit stuck on wraping my head around all the new concepts (I just finished a beginners course covering material on elementary number theory, algebraic structures of the integers, some basic abstract algebra - concerning only finite and abelian groups -, with hints on analytic number theory and deeper abstract algebra). I read the start of Serre's 1978 lectures Groupes Finis as a motivation, then moved on to Bourbaki's Algebra.
Bourbaki's encyclopedia was not hard to follow up to chapter 6, where extensions began popping up. The characterization of a extension was comprehensible, along with his demonstration that the direct product and the external direct product were extensions and their characterizations. But when he moved on to the (omissed term internal) semidirect product, I felt unsure on my understanding of the subject. Here's what I've got:
Bourbaki's definition, external semidirect product: If $G$ and $F$ are two groups and there exists a homomorphism $\pi$ of $G$ into $\text{Aut}(F)$ - (the automorphism group of F, which in my understanding clearly defines a action of $G$ on $F$, through the map $\phi :G\times F \rightarrow F$ where $\phi(g, f) \mapsto \pi(g)(f)$) - then the set $F \times G$ with the composition:
$$((f, g), (f',g')) \mapsto (f\pi(g)(f'), gg')$$
is a extension of G by F (written $F \times_\pi G$). With that proposition, we can describe all external semidirect product extensions of two groups $F$ and $G$ by looking at the possible homomorphisms of the automorphism group.
Now to the real question. Bourbaki's definition, semidirect product: $G$ is a semidirect product of the subgroups $H$ and $K$ if $H$ is a normal subgroup, if $K\cap H = \{e\}$ and $H.K = G$. First off, from my understanding the statement $H.K = G$ is referring to the property of complement subgroups, i.e., every element $g$ of $G$ can be uniquely written as $hk$ for some $h \in H$ and $k \in K$. That means $K$ has one and only one representative of each orbit of the equivalence relation defined by $H$, and that means $K$ is isomorphic to $G/H$. Now, Bourbaki proceeds: "Let $\pi$ be the operation of $K$ on $H$ by inner automorphisms of $G$. The mapping $(h, k) \mapsto hk$ is a isomorphism of $H \times_\pi K$ onto $G$".
I'm not sure I actually understand that last $\pi$ operation. If it defines a external semidirect product, it must be a homomorphism from $K$ to the automorphism group of $H$, so it must define a action of $K$ on $H$. When he mentions inner automorphisms of $G$, because $H$ is normal every inner automorphism of $G$ is a automorphism of $H$, and I guess the converse is also valid, so why couldn't $\pi$ be defined in terms of the group of automorphisms of $H$? Would it make it different? Does the set of inner automorphisms of $G$ and automorphisms of $H$ differ?
 A: The reference to "inner automorphisms" here is just to say that $K$ is acting on $H$ by conjugation.  That is, the homomorphism $\pi:K\to\operatorname{Aut}(H)$ is defined by sending $k$ to the automorphism $h\mapsto khk^{-1}$ (this maps $H$ to itself because $H$ is normal in $G$).  Note that you shouldn't say $K$ is acting on $H$ by inner automorphisms of $H$, because such an automorphism might not be an inner automorphism of $H$.  For something to be an inner automorphism of $H$, it must be of the form $h\mapsto khk^{-1}$ for some element $k$ of $H$, whereas here our $k$ is in $K$, not $H$.
(Also, automorphisms of $H$ and inner automorphisms of $G$ might be different; there is no reason that every automorphism of $H$ must be given by conjugation by some element of $G$, and there might be elements of $G$ which give different inner automorphisms on all of $G$ but which restrict to the same map on $H$.  All you can say in general is that there is a homomorphism from the inner automorphisms of $G$ to the automorphisms of $H$, given by restriction.)
