Holomorph (mathematics)

If ${\displaystyle \operatorname {Aut} (G)}$ is the automorphism group of ${\displaystyle G}$, then

${\displaystyle \operatorname {Hol} (G)=G\rtimes \operatorname {Aut} (G)}$,

where the multiplication is given by

${\displaystyle (g,\alpha )(h,\beta )=(g\alpha (h),\alpha \beta ).}$

1

Typically, a semidirect product is given in the form ${\displaystyle G\rtimes _{\phi }A}$, where ${\displaystyle G}$ and ${\displaystyle A}$ are groups and ${\displaystyle \phi :A\rightarrow \operatorname {Aut} (G)}$ is a homomorphism, and where the multiplication of elements in the semidirect product is given as

${\displaystyle (g,a)(h,b)=(g\phi (a)(h),ab)}$.

This is well defined since ${\displaystyle \phi (a)\in \operatorname {Aut} (G)}$, and therefore ${\displaystyle \phi (a)(h)\in G}$.

For the holomorph, ${\displaystyle A=\operatorname {Aut} (G)}$ and ${\displaystyle \phi }$ is the identity map. As such, we suppress writing ${\displaystyle \phi }$ explicitly in the multiplication given in equation (1) above.

As an example, take

• ${\displaystyle G=C_{3}=\langle x\rangle =\{1,x,x^{2}\}}$ the cyclic group of order 3,
• ${\displaystyle \operatorname {Aut} (G)=\langle \sigma \rangle =\{1,\sigma \}}$, where ${\displaystyle \sigma (x)=x^{2}}$, and
• ${\displaystyle \operatorname {Hol} (G)=\{(x^{i},\sigma ^{j})\}}$ with the multiplication given by:

${\displaystyle (x^{i_{1}},\sigma ^{j_{1}})(x^{i_{2}},\sigma ^{j_{2}})=(x^{i_{1}+i_{2}2^{^{j_{1}}}},\sigma ^{j_{1}+j_{2}})}$, where the exponents of ${\displaystyle x}$ are taken mod 3 and those of ${\displaystyle \sigma }$ mod 2.

Observe that

${\displaystyle (x,\sigma )(x^{2},\sigma )=(x^{1+2\cdot 2},\sigma ^{2})=(x^{2},1)}$ while ${\displaystyle (x^{2},\sigma )(x,\sigma )=(x^{2+1\cdot 2},\sigma ^{2})=(x,1)}$.

Hence, this group is not abelian, and so ${\displaystyle \operatorname {Hol} (C_{3})}$ is a non-abelian group of order 6, which, by basic group theory, must be isomorphic to the symmetric group ${\displaystyle S_{3}}$.