# example of infinite simple group

This fact that finite alternating groups^{} are simple can be extended to a result about an infinite group. Let $G$ be the subgroup^{} of the group of permutations^{} on a countably infinite^{} set $M$ (which we may take to be the set of natural numbers for concreteness) which is generated by cycles of length $3$. Note that any since every element of this group is a product^{} of a finite number of cycles, the permutations of $G$ are such that only a finite number of elements of our set are not mapped to themselves by a given permutation.

We will now show that $G$ is simple. Suppose that $\pi $ is an element of $G$ other than the identity^{}. Let $m$ be the set of all $x$ such that $\pi (x)\ne x$. By our previous comment, $m$ is finite. Consider the restriction^{} ${\pi}_{m}$ of $\pi $ to $m$. By the theorem of the parent entry (http://planetmath.org/SimplicityOfA_n), the subgroup of ${A}_{m}$ generated by the conjugates of ${\pi}_{m}$ is the whole of ${A}_{m}$. In particular, this means that there exists a cycle of order $3$ in ${A}_{m}$ which can be expressed as a product of ${\pi}_{m}$ and its conjugates. Hence the subgroup of $G$ generated by conjugates of $\pi $ contains a cycle of length three as well. However, every cycle of order $3$ is conjugate to every other cycle of order $3$ so, in fact, the subgroup of $G$ generated by the conjugates of $\pi $ is the whole of $G$. Hence, the only normal subgroups^{} of $G$ are the group consisting of solely the identity element^{} and the whole of $G$, so $G$ is a simple group^{}.

Title | example of infinite^{} simple group |
---|---|

Canonical name | ExampleOfInfiniteSimpleGroup |

Date of creation | 2013-03-22 16:53:31 |

Last modified on | 2013-03-22 16:53:31 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Example |

Classification | msc 20E32 |

Classification | msc 20D06 |