# Semi-direct product

of a group $A$ by a group $B$
A group $G = AB$ which is the product of its subgroups $A$ and $B$, where $B$ is normal in $G$ and $A \cap B = \{1\}$. If $A$ is also normal in $G$, then the semi-direct product becomes a direct product. The semi-direct product of two groups $A$ and $B$ is not uniquely determined. To construct a semi-direct product one should also know which automorphisms of the group $B$ are induced by conjugation by elements of $A$. More precisely, if $G = AB$ is a semi-direct product, then to each element $a \in A$ corresponds an automorphism $\alpha_a \in \mathrm{Aut}(B)$, which is conjugation by the element $a$: $$\alpha_a(b) = a b a^{-1}\,,\ \ \ b \in B \ .$$ Here, the correspondence $a \mapsto \alpha_a$ is a homomorphism $A \rightarrow \mathrm{Aut}(B)$. Conversely, if $A$ and $B$ are arbitrary groups, then for any homomorphism $\phi : A \rightarrow \mathrm{Aut}(B)$ there is a unique semi-direct product of the group $A$ by the group $B$ for which $\alpha_a = \phi(a)$ for any $a \in A$. A semi-direct product is a particular case of an extension of a group $B$ by a group $A$ (cf. Extension of a group); such an extension is called split.