# the derived subgroup is normal

We are going to prove:

”The derived subgroup (or commutator subgroup) $[G,G]$ is normal in $G$”

Proof:

We have to show that for each $x\in [G,G]$, $gx{g}^{-1}$ it is also in $[G,G]$.

Since $[G,G]$ is the subgroup^{} generated by the all commutators in $G$, then for each $x\in [G,G]$ we have $x={c}_{1}{c}_{2}\mathrm{\cdots}{c}_{m}$ –a word of commutators– so ${c}_{i}=[{a}_{i},{b}_{i}]$ for all $i$.

Now taking any element of $g\in G$ we can see that

$g[{a}_{i},{b}_{i}]{g}^{-1}$ | $=$ | $g{a}_{i}{b}_{i}{a}_{i}^{-1}{b}_{i}^{-1}{g}^{-1}$ | ||

$=$ | $g{a}_{i}{g}^{-1}g{b}_{i}{g}^{-1}g{a}_{i}^{-1}{g}^{-1}g{b}_{i}^{-1}{g}^{-1}$ | |||

$=$ | $(g{a}_{i}{g}^{-1})(g{b}_{i}{g}^{-1}){(g{a}_{i}{g}^{-1})}^{-1}{(g{b}_{i}{g}^{-1})}^{-1}$ | |||

$=$ | $[g{a}_{i}{g}^{-1},g{b}_{i}{g}^{-1}],$ |

that is

$$g[{a}_{i},{b}_{i}]{g}^{-1}=[g{a}_{i}{g}^{-1},g{b}_{i}{g}^{-1}]$$ |

so a conjugation^{} of a commutator is another commutator, then
for the conjugation

$gx{g}^{-1}$ | $=$ | $g{c}_{1}{c}_{2}\mathrm{\cdots}{c}_{m}{g}^{-1}$ | ||

$=$ | $g{c}_{1}{g}^{-1}g{c}_{2}{g}^{-1}g\mathrm{\cdots}{g}^{-1}g{c}_{m}{g}^{-1}$ | |||

$=$ | $(g{c}_{1}{g}^{-1})(g{c}_{2}{g}^{-1})\mathrm{\cdots}(g{c}_{m}{g}^{-1})$ |

is another word of commutators, hence $gx{g}^{-1}$ is in $[G,G]$ which in turn implies that $[G,G]$ is normal in $G$, QED.

Title | the derived subgroup is normal |
---|---|

Canonical name | TheDerivedSubgroupIsNormal |

Date of creation | 2013-03-22 16:04:39 |

Last modified on | 2013-03-22 16:04:39 |

Owner | juanman (12619) |

Last modified by | juanman (12619) |

Numerical id | 8 |

Author | juanman (12619) |

Entry type | Proof |

Classification | msc 20A05 |

Classification | msc 20E15 |

Classification | msc 20F14 |