# finite subgroup

Theorem. A non-empty finite subset $K$ of a group $G$ is a subgroup^{} of $G$ if and only if

$xy\in K\mathit{\hspace{1em}}\text{for all}\mathit{\hspace{1em}}x,y\in K.$ | (1) |

*Proof.* The condition (1) is apparently true if $K$ is a subgroup. Conversely, suppose that a nonempty finite subset $K$ of the group $G$ satisfies (1). Let $a$ and $b$ be arbitrary elements of $K$. By (1), all () powers of $b$ belong to $K$. Because of the finiteness of $K$, there exist positive integers $r,s$ such that

$${b}^{r}={b}^{s},r>s+1.$$ |

By (1),

$$K\ni {b}^{r-s-1}={b}^{r-s}{b}^{-1}=e{b}^{-1}={b}^{-1}.$$ |

Thus also $a{b}^{-1}\in K$, whence, by the theorem of the http://planetmath.org/node/1045parent entry, $K$ is a subgroup of $G$.

Example. The multiplicative group^{} $G$ of all nonzero complex numbers^{} has the finite multiplicative subset
$\{1,-1,i,-i\}$, which has to be a subgroup of $G$.

Title | finite subgroup |
---|---|

Canonical name | FiniteSubgroup |

Date of creation | 2013-03-22 18:57:02 |

Last modified on | 2013-03-22 18:57:02 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 20A05 |

Synonym | criterion for finite subgroup |

Synonym | finite subgroup criterion |