example of infinite simple group
This fact that finite alternating groups are simple can be extended to a result about an infinite group. Let be the subgroup of the group of permutations on a countably infinite set (which we may take to be the set of natural numbers for concreteness) which is generated by cycles of length . Note that any since every element of this group is a product of a finite number of cycles, the permutations of 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 is simple. Suppose that is an element of other than the identity. Let be the set of all such that . By our previous comment, is finite. Consider the restriction of to . By the theorem of the parent entry (http://planetmath.org/SimplicityOfA_n), the subgroup of generated by the conjugates of is the whole of . In particular, this means that there exists a cycle of order in which can be expressed as a product of and its conjugates. Hence the subgroup of generated by conjugates of contains a cycle of length three as well. However, every cycle of order is conjugate to every other cycle of order so, in fact, the subgroup of generated by the conjugates of is the whole of . Hence, the only normal subgroups of are the group consisting of solely the identity element and the whole of , so is a simple group.
|Title||example of infinite simple group|
|Date of creation||2013-03-22 16:53:31|
|Last modified on||2013-03-22 16:53:31|
|Last modified by||rspuzio (6075)|