Holomorph (mathematics)

Hol(''G'') as a semidirect product

If \operatorname{Aut}(G) is the automorphism group of G, then :\operatorname{Hol}(G)=G\rtimes \operatorname{Aut}(G), where the multiplication is given by

Typically, a semidirect product is given in the form G\rtimes_{\phi}A, where G and A are groups and \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\phi(a)(h),ab). This is well defined since \phi(a)\in \operatorname{Aut}(G), and therefore \phi(a)(h)\in G.

For the holomorph, A=\operatorname{Aut}(G) and \phi is the identity map. As such, we suppress writing \phi explicitly in the multiplication given in equation () above.

As an example, take

• G=C_3=\langle x\rangle={1,x,x^2} the cyclic group of order 3,
• \operatorname{Aut}(G)=\langle \sigma\rangle={1,\sigma}, where \sigma(x)=x^2, and
• \operatorname{Hol}(G)={(x^i,\sigma^j)} with the multiplication given by: :(x^{i_1},\sigma^{j_1})(x^{i_2},\sigma^{j_2}) = (x^{i_1+i_22^{^{j_1}}},\sigma^{j_1+j_2}), where the exponents of x are taken mod 3 and those of \sigma mod 2.

Observe that :(x,\sigma)(x^2,\sigma)=(x^{1+2\cdot2},\sigma^2)=(x^2,1) while (x^2,\sigma)(x,\sigma)=(x^{2+1\cdot2},\sigma^2)=(x,1). Hence, this group is not abelian, and so \operatorname{Hol}(C_3) is a non-abelian group of order 6, which, by basic group theory, must be isomorphic to the symmetric group S_3.

Hol(''G'') as a permutation group

A group G acts naturally on itself by left and right multiplication, each giving rise to a homomorphism from G into the symmetric group on the underlying set of G. One homomorphism is defined as λ: G → Sym(G), λg(h) = g·h. That is, g is mapped to the permutation obtained by left-multiplying each element of G by g. Similarly, a second homomorphism ρ: G → Sym(G) is defined by ρg(h) = h·g−1, where the inverse ensures that ρgh(k) = ρg(ρh(k)). These homomorphisms are called the left and right regular representations of G. Each homomorphism is injective, a fact referred to as Cayley's theorem.

For example, if G = C3 = {1, x, x2 } is a cyclic group of order three, then

• λx(1) = x·1 = x,
• λx(x) = x·x = x2, and
• λx(x2) = x·x2 = 1, so λ(x) takes (1, x, x2) to (x, x2, 1).

The image of λ is a subgroup of Sym(G) isomorphic to G, and its normalizer in Sym(G) is defined to be the holomorph N of G. For each n in N and g in G, there is an h in G such that n·λg = λh·n. If an element n of the holomorph fixes the identity of G, then for 1 in G, (n·λg)(1) = (λh·n)(1), but the left hand side is n(g), and the right side is h. In other words, if n in N fixes the identity of G, then for every g in G, n·λg = λn(g)·n. If g, h are elements of G, and n is an element of N fixing the identity of G, then applying this equality twice to n·λg·λh and once to the (equivalent) expression n·λgg gives that n(g)·n(h) = n(g·h). That is, every element of N that fixes the identity of G is in fact an automorphism of G. Such an n normalizes λG, and the only λg that fixes the identity is λ(1). Setting A to be the stabilizer of the identity, the subgroup generated by A and λG is semidirect product with normal subgroup λG and complement A. Since λG is transitive, the subgroup generated by λG and the point stabilizer A is all of N, which shows the holomorph as a permutation group is isomorphic to the holomorph as semidirect product.

It is useful, but not directly relevant, that the centralizer of λG in Sym(G) is ρG, their intersection is \rho_{Z(G)}=\lambda_{Z(G)}, where Z(G) is the center of G, and that A is a common complement to both of these normal subgroups of N.

Properties

• ρ(G) ∩ Aut(G) = 1
• Aut(G) normalizes ρ(G) so that canonically ρ(G)Aut(G) ≅ G ⋊ Aut(G)
• \operatorname{Inn}(G)\cong \operatorname{Im}(g\mapsto \lambda(g)\rho(g)) since λ(g)ρ(g)(h) = ghg−1 (\operatorname{Inn}(G) is the group of inner automorphisms of G.)
• K ≤ G is a characteristic subgroup if and only if λ(K) ⊴ Hol(G)

References