In mathematics, especially in the area of algebra known as group theory, the holomorph of a group G {\displaystyle G} , denoted Hol ( G ) {\displaystyle \operatorname {Hol} (G)} , is a group that simultaneously contains (copies of) G {\displaystyle G} and its automorphism group Aut ( G ) {\displaystyle \operatorname {Aut} (G)} . It provides interesting examples of groups, and allows one to treat group elements and group automorphisms in a uniform context. The holomorph can be described as a semidirect product or as a permutation group.
Hol(G) as a semidirect product If Aut ( G ) {\displaystyle \operatorname {Aut} (G)} is the automorphism group of G {\displaystyle G} , then
Hol ( G ) = G ⋊ Aut ( G ) {\displaystyle \operatorname {Hol} (G)=G\rtimes \operatorname {Aut} (G)} , where the multiplication is given by
Typically, a semidirect product is given in the form G ⋊ ϕ A {\displaystyle G\rtimes _{\phi }A} , where G {\displaystyle G} and A {\displaystyle A} are groups and ϕ : A → Aut ( G ) {\displaystyle \phi :A\rightarrow \operatorname {Aut} (G)} is a homomorphism, and where the multiplication of elements in the semidirect product is given as
( g , a ) ( h , b ) = ( g ϕ ( a ) ( h ) , a b ) {\displaystyle (g,a)(h,b)=(g\phi (a)(h),ab)} . This is well defined since ϕ ( a ) ∈ Aut ( G ) {\displaystyle \phi (a)\in \operatorname {Aut} (G)} , and therefore ϕ ( a ) ( h ) ∈ G {\displaystyle \phi (a)(h)\in G} . For the holomorph, A = Aut ( G ) {\displaystyle A=\operatorname {Aut} (G)} and ϕ {\displaystyle \phi } is the identity map. As such, we suppress writing ϕ {\displaystyle \phi } explicitly in the multiplication given in equation (1) above. As an example, take
G = C 3 = ⟨ x ⟩ = { 1 , x , x 2 } {\displaystyle G=C_{3}=\langle x\rangle =\{1,x,x^{2}\}} the cyclic group of order 3,
Aut ( G ) = ⟨ σ ⟩ = { 1 , σ } {\displaystyle \operatorname {Aut} (G)=\langle \sigma \rangle =\{1,\sigma \}} , where σ ( x ) = x 2 {\displaystyle \sigma (x)=x^{2}} , and
Hol ( G ) = { ( x i , σ j ) } {\displaystyle \operatorname {Hol} (G)=\{(x^{i},\sigma ^{j})\}} with the multiplication given by:
( x i 1 , σ j 1 ) ( x i 2 , σ j 2 ) = ( x i 1 + i 2 2 j 1 , σ j 1 + j 2 ) {\displaystyle (x^{i_{1}},\sigma ^{j_{1}})(x^{i_{2}},\sigma ^{j_{2}})=(x^{i_{1}+i_{2}2^{^{j_{1}}}},\sigma ^{j_{1}+j_{2}})} , where the exponents of x {\displaystyle x} are taken mod 3 and those of σ {\displaystyle \sigma } mod 2. Observe that
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