How does action by conjugation determine the product stucture of a semidirect product? Consider $G=P\ltimes Q$ where $P\cap Q=\{e\}$ and $Q<N_G(P)$. Here $\ltimes$ is the inner semidirect product.
Here I believe it is the case that the conjugation action of $P$ on $Q$ will determine the product structure of $G$.
So how do we know that the product structure on $G$ is determined by the conjugation action of $P$ on $Q$?
Is it correct to say that the automorphisms of $G$ determine the possible product structures, and so if we can construct a surjective map from the possible conjugation actions of $P$ on $Q$ to the automorphisms of $G$ then we will know that the conjugation action determines the product structure of $G$?
If so, what notation would express that map generally?
Can a concrete map be demonstrated for this case without referring to external semidirect products? I think that there's a generalization of this involving external semidirect products, but I'm trying to work up to that.
The context here is that I'm studying a classification of groups of order 6. I understand that classification pretty well in the concrete sense, but am trying to really get my head around the abstract ideas before moving on to harder examples.
 A: Not sure if this is what you mean. For given $P,Q$, the semidirect group notation $P\ltimes Q$ is ambiguous unless you specify an action of $P$ on $Q$. (However, you wil often find the definition of the action omitted if it is "understood" or there is only one natural choice for a nontrivial action). Once you have that, $P, Q$ become subgroups of $P\ltimes Q$ (more formally: there exists a canonical embedding) and these subgroups have the property that $Q$ is normal and the original action of $P$ on $Q$ turns into conjugation. For the inner semidirect product we start at this end, i.e. the "secret" actoin of $P$ on $Q$ is already conjugation. Since $P,Q$ also generate $G$, multiplication in $G$ is indeed determined by this conjugation.
A priori, "$P,Q$ generate $G$" merely means that each element of $G$ can be written as $p_1q_1p_2q\cdots p_nq_n$ with $p_i\in P, q_i\in Q$. If you know how to conjugate, you can shorten such products: If $n>1$, there is $p'$ with $q_{n-1}p_n=p'q_{n-1}$ and then replacing $p_{n-1}p'$ and $q_{n-1}q_n$ with single elements of $P$, $Q$ we obtain a shorter product. Hence ultimatley all elements of $G$ can be written in the form $pq$ and their multiplication $p_1q_1\cdot p_2q_2=p_3q_3$ is again governed by the permutation action.
