Difference between revisions of "Semi-direct product"
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''of a group $A$ by a group $B$'' | ''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 $a$: | + | 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 $a$: |
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\alpha_a(b) = a b a^{-1}\,,\ \ \ b \in B \ . | \alpha_a(b) = a b a^{-1}\,,\ \ \ b \in B \ . | ||
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====References==== | ====References==== | ||
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− | <TR><TD valign="top">[ | + | <TR><TD valign="top">[a1]</TD> <TD valign="top"> Paul M. Cohn. ''Basic Algebra: Groups, Rings, and Fields'', Springer (2003) {{ISBN|1852335874}} {{ZBL|1003.00001}}</TD></TR> |
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Latest revision as of 16:54, 23 November 2023
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.
References
[1] | A.G. Kurosh, "The theory of groups" , 1 , Chelsea (1960) (Translated from Russian) |
Comments
The semi-direct product of $A$ by $B$ is often denoted by $B \rtimes A$ or $B : A$.
The term "internal" semi-direct product is used for the case when $A$ and $B$ are considered as subgroups of the given group $G$. The "external" semi-direct product of groups $A$ and $B$, with a map $\phi : A \rightarrow \mathrm{Aut}(B)$ , may be taken to be the Cartesian product $A \times B$ with multiplication defined by $$ (a_1,b_1) \cdot (a_2,b_2) = \left({a_1a_2, b_1^{\phi(a_2)}b_2}\right) \ . $$
References
[a1] | Paul M. Cohn. Basic Algebra: Groups, Rings, and Fields, Springer (2003) ISBN 1852335874 Zbl 1003.00001 |
Semi-direct product. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Semi-direct_product&oldid=33934