simplicity of the alternating groups
Theorem 1.
If , then the alternating group
on symbols, , is simple.
Throughout the proof we extensively employ the cycle notation, with
composition on the left, as is usual. The symmetric group
on
symbols is denoted by .
The following observation will be useful. Let be a permutation
written as disjoint cycles
It is easy to check that for every permutation we have
As a consequence, two permutations of are conjugate exactly when
they have the same cycle type.
Two preliminary results will also be necessary.
Lemma 2.
The set of cycles of length generates .
Proof.
A product of -cycles is an even permutation
, so the subgroup
generated by all -cycles is therefore contained in . For the reverse inclusion, by definition every even permutation is the product of even number
of transpositions
. Thus, it suffices to show that the product of two transpositions can be written as a product of -cycles. There are two possibilities. Either the two transpositions move an element in common, say and , or the two transpositions are disjoint, say and . In the former case,
and in the latter,
This establishes the first lemma. ∎
Lemma 3.
If a normal subgroup contains a -cycle, then
.
Proof.
We will show that if , then the assumption of normality implies that any other . This is easy to show, because there is some permutation in that under conjugation
takes to , that is
In case is odd, then (because ) we can choose some transposition disjoint from so that
that is,
where is even. This means that contains all -cycles, as . Hence, by previous lemma as required. ∎
Proof of theorem.
Let be a non-trivial normal subgroup. We will show that . The proof now proceeds by cases. In each case, the normality of will allow us to reduce the proof to Lemma 2 or to one of the previous cases.
Case 1.
Suppose that there exists a that, when written as disjoint cycles, has a cycle of length at least , say
Upon conjugation by , we obtain
Hence, , and hence also. Notice that the rest of the cycles cancel. By Lemma 3, .
Case 2.
Suppose that there exists a whose disjoint cycle decomposition has at least two cycles of length 3, say
Conjugation by implies that also contains
Hence, also contains . This reduces the proof to Case 1.
Case 3.
Case 4.
Suppose there exists a of the form . Conjugating by with distinct from (at least one such exists, as ) yields
Hence . Again, Lemma 3 applies.
Case 5.
Suppose that contains a permutation of the form
This time we conjugate by .
Observe that
which reduces the proof to Case 2.
Since there exists at least one non-identity , and since this is covered by one of the above cases, we conclude that , as was to be shown.
Title | simplicity of the alternating groups |
---|---|
Canonical name | SimplicityOfTheAlternatingGroups |
Date of creation | 2013-03-22 13:07:57 |
Last modified on | 2013-03-22 13:07:57 |
Owner | rmilson (146) |
Last modified by | rmilson (146) |
Numerical id | 16 |
Author | rmilson (146) |
Entry type | Result |
Classification | msc 20D06 |
Classification | msc 20E32 |
Related topic | ExamplesOfFiniteSimpleGroups |