# semi-direct factor and quotient group

###### Theorem.

If the group $G$ is a semi-direct product of its subgroups^{} $H$ and $Q$,
then the semi-direct $Q$
is isomorphic^{} to the quotient group^{} $G/H$.

Proof. Every element $g$ of $G$ has the unique representation $g=hq$ with $h\in H$ and $q\in Q$. We therefore can define the mapping

$$g\mapsto q$$ |

from $G$ to $Q$.
The mapping is surjective^{} since any element $y$ of $Q$ is the image of $ey$.
The mapping is also a homomorphism^{} since if ${g}_{1}={h}_{1}{q}_{1}$ and ${g}_{2}={h}_{2}{q}_{2}$, then we obtain

$$f({g}_{1}{g}_{2})=f({h}_{1}{q}_{1}{h}_{2}{q}_{2})=f({h}_{1}{h}_{2}{q}_{1}{q}_{2})={q}_{1}{q}_{2}=f({g}_{1})f({g}_{2}).$$ |

Then we see that $\mathrm{ker}f=H$ because all elements $h=he$ of $H$
are mapped to the identity element^{} $e$ of $Q$.
Consequently we get, according to the first isomorphism theorem^{}, the result

$$G/H\cong Q.$$ |

Example.
The multiplicative group^{} ${\mathbb{R}}^{\times}$ of reals
is the semi-direct product of the subgroups
$\{1,-1\}=\{\pm 1\}$ and ${\mathbb{R}}_{+}$.
The quotient group ${\mathbb{R}}^{\times}/\{\pm 1\}$ consists of all cosets

$$x\{\pm 1\}=\{x,-x\}$$ |

where $x\ne 0$, and is obviously isomorphic with ${\mathbb{R}}_{+}=\{x\mid x>0\}$.

Title | semi-direct factor and quotient group |
---|---|

Canonical name | SemidirectFactorAndQuotientGroup |

Date of creation | 2013-03-22 15:10:22 |

Last modified on | 2013-03-22 15:10:22 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 8 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 20E22 |