# example of fully invariant subgroup

The derived subgroup $[G,G]$ is a fully invariant subgroup because if $f$ is an endomorphism^{} of $G$, then for each word of commutators^{} $[{a}_{1},{b}_{1}][{a}_{2},{b}_{2}]\mathrm{\cdots}[{a}_{m},{b}_{m}]$, we have

$$f([{a}_{1},{b}_{1}][{a}_{2},{b}_{2}]\mathrm{\cdots}[{a}_{m},{b}_{m}])=[f{a}_{1},f{b}_{1}][f{a}_{2},f{b}_{2}]\mathrm{\cdots}[f{a}_{m},f{b}_{m}]\in [G,G]$$ |

i.e. the homomorphic image^{} of a word of commutators is a word of commutators.

Title | example of fully invariant subgroup |
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Canonical name | ExampleOfFullyInvariantSubgroup |

Date of creation | 2013-03-22 16:04:41 |

Last modified on | 2013-03-22 16:04:41 |

Owner | juanman (12619) |

Last modified by | juanman (12619) |

Numerical id | 5 |

Author | juanman (12619) |

Entry type | Example |

Classification | msc 20D99 |

Related topic | FullyInvariantSubgroup |