Is $\langle A,G\rangle=G\rtimes_{\sigma}A$? Given groups $A$ and $G$; suppose $A$ act via automorphism on $G$, i.e. there exists $\sigma:A\rightarrow\operatorname{Aut}G$, $a\mapsto\sigma_a:g\mapsto g^a$.
From this action we can define a semidirect product $G\rtimes_{\sigma}A$, which is a group that contains (isomorphic copies of) $A$ and $G$.
Then we know by definition that $\langle A,G\rangle$ is the smallest group containing both $A$ and $G$. Hence, up to isomorphism, we have that $\langle A,G\rangle\le G\rtimes_{\sigma}A$.
Is the equality always true? I think yes in the case $A\cap G=1$, but in general?
 A: Insofar as your question makes sense, yes, it is true. $A \cap G=1$ by definition. If you are confused about that, then that is fine, because the question itself is confused. $\langle A,G\rangle$ only makes sense when $A$ and $G$ are subgroups of some larger group. Once you pick that larger group $X$, then $\langle A,G \rangle$ is a subgroup of $X$. If you pick $X=G \rtimes A$, then by definition of $\rtimes$, $A \cap G =1$.
However, depending on the groups $A$ and $G$, it could be possible to pick different $X$. These could have different values of $A \cap G$, and different values of $\langle A,G\rangle$.
All we know about $X$ (before you say that it is the semi-direct product) is that $A,G \leq X$ and $A \leq N_X(G)$. Inside any such $X$, we have $\langle A,G\rangle$ is a quotient of the semi-direct product.
It is standard when talking about groups acting on other groups to take $X$ to be the semi-direct product. This produces the largest possible value of $\langle A,G\rangle$, and usually the only results we care about are those that hold not only in $G \rtimes A$, but also in every quotient, but those results are phrased in terms of the semi-direct product because the corresponding results in quotients either have more awkward phrasings or identical phrasings, so are either hard to state generally (but easy to handle in any particular case) or so easy that there is no need to state it.
