example of infinite simple group

This fact that finite alternating groupsMathworldPlanetmath are simple can be extended to a result about an infinite group. Let G be the subgroupMathworldPlanetmathPlanetmath of the group of permutationsMathworldPlanetmath on a countably infiniteMathworldPlanetmath 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 productPlanetmathPlanetmath 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 π is an element of G other than the identityPlanetmathPlanetmathPlanetmath. Let m be the set of all x such that π(x)x. By our previous comment, m is finite. Consider the restrictionPlanetmathPlanetmathPlanetmath πm of π to m. By the theorem of the parent entry (http://planetmath.org/SimplicityOfA_n), the subgroup of Am generated by the conjugates of πm is the whole of Am. In particular, this means that there exists a cycle of order 3 in Am which can be expressed as a product of πm and its conjugates. Hence the subgroup of G generated by conjugates of π 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 π is the whole of G. Hence, the only normal subgroupsMathworldPlanetmath of G are the group consisting of solely the identity elementMathworldPlanetmath and the whole of G, so G is a simple groupMathworldPlanetmathPlanetmath.

Title example of infiniteMathworldPlanetmathPlanetmath 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