Classification of ambivalent alternating groups: Difference between revisions

Revision as of 01:59, 30 October 2010

Statement

The Alternating group (?) $A_{n}$ is an Ambivalent group (?) for precisely the following choices of $n$ : $n=1,2,5,6,10,14$ .

Note that in the proof, we show that in each of these cases, for every element in the alternating group, there is an element of order two in the alternating group conjugating it to its inverse. Thus, we show that the set of $n$ for which $A_{n}$ is strongly ambivalent is precisely the same: $n=1,2,5,6,10,14$ .

Related facts

Related facts about alternating groups

Classification of alternating groups having a class-inverting automorphism: This turns out to be $n=1,2,3,4,5,6,7,8,10,12,14$ .
Alternating group implies every element is automorphic to its inverse
Finitary alternating group on infinite set is ambivalent

Related facts about symmetric groups

General information pages

Facts used

Criterion for element of alternating group to be real

Proof

By fact (1), a product of cycles of distinct odd lengths $r_{1},r_{2},\dots ,r_{k}$ is conjugate to its inverse if and only if $\sum (r_{i}-1)/2$ is even. Equivalently, it is conjugate to its inverse if and only if the number of $r_{i}$ s that are congruent to $3$ modulo $4$ is even. Note also that if it is conjugate to its inverse, we can choose as our conjugating element an element of order two: the product of transpositions described above.

What this boils down to for $n$

Thus, the problem reduces to the following: for what $n$ can we write $n=\sum _{i=1}^{k}r_{i}$ in such a way that all $r_{i}$ are distinct, and the number of $r_{i}$ that are congruent to $3$ modulo $4$ is odd? These are precisely the $n$ for which $A_{n}$ is not ambivalent.

We quickly see the following:

$n=4d+3$ can be written in this form, because we can take $k=1,r_{1}=4d+3$ .
$n=4d+4$ can be written in this form, because we can take $k=2,r_{1}=4d+3,r_{2}=1$ .
$n=4d+9$ can be written in this form, because we can take $k=3,r_{1}=1,r_{2}=5,r_{3}=4d+3$ .
$n=4d+18$ can be written in this form, because we can take $k=4,r_{1}=1,r_{2}=5,r_{3}=9,r_{4}=4d+3$ .

The only cases left are $n=1,2,5,6,10,14$ , and it is readily seen that a decomposition into $r_{i}$ of the above form is not possible for these $n$ .

@@ Line 24: / Line 24: @@
 ==Facts used==
-# [[uses::Splitting criterion for conjugacy classes in the alternating group]]
+# [[uses::Criterion for element of alternating group to be real]]
 ==Proof==
-===Overall plan===
+By fact (1), a product of cycles of distinct odd lengths <math>r_1,r_2,\dots,r_k</math> is conjugate to its inverse if and only if <math>\sum (r_i - 1)/2</math> is even. Equivalently, it is conjugate to its inverse if and only if the number of <math>r_i</math>s that are congruent to <math>3</math> modulo <math>4</math> is even. Note also that if it is conjugate to its inverse, we can choose as our conjugating element an element of order two: the product of transpositions described above.
-If <math>g \in A_n</math> has the property that its conjugacy class in <math>S_n</math> does ''not'' split in <math>A_n</math>, then <math>g</math> is conjugate to <math>g^{-1}</math> in <math>A_n</math> (because they're conjugate in <math>S_n</math>). Thus, it suffices to check whether every element whose conjugacy class ''does'' split inside <math>A_n</math>, is conjugate to its inverse.
-By the splitting criterion for conjugacy classes, it suffices to look at those even permutations that arise as products of cycles of distinct odd lengths. Further, in order to determine whether such a product of cycles is conjugate to its inverse in <math>A_n</math>, it suffices to find ''one'' permutation that conjugates this cycle to its inverse. If that one permutation is even, then the element is conjugate to its inverse. If that one permutation is odd, then the element is not conjugate to its inverse in <math>A_n</math>.
-===Criterion for determining whether an element is conjugate to its inverse===
-For a cycle of odd length <math>(a_1, a_2, \dots, a_r)</math>, the product of <math>(r-1)/2</math> transpositions <math>(a_j,a_{r+2-j})</math> conjugates this cycle to its inverse. Thus, if a permutation is a product of cycles of odd lengths <math>r_1,r_2,\dots,r_k</math>, then there is a product of <math>\sum (r_i - 1)/2</math> transpositions that conjugates this to its inverse.
-The upshot: a product of cycles of distinct odd lengths <math>r_1,r_2,\dots,r_k</math> is conjugate to its inverse if and only if <math>\sum (r_i - 1)/2</math> is even. Equivalently, it is conjugate to its inverse if and only if the number of <math>r_i</math>s that are congruent to <math>3</math> modulo <math>4</math> is even. Note also that if it is conjugate to its inverse, we can choose as our conjugating element an element of order two: the product of transpositions described above.
 ===What this boils down to for <math>n</math>===
-Thus, the problem reduces to the following: for what <math>n</math> can we write <math>n = \sum_{i=1}^k r_i</math> in such a way that all <math>r_i</math> are distinct, and the number of <math>r_i</math> that are congruent to <math>3</math> modulo <math>4</math> is odd? These are precisely the <math>n</math> for which <math>A_n</math> is not ambivalent.
+Thus, the problem reduces to the following: for what <math>n</math> can we write <math>n = \sum_{i=1}^k r_i</math> in such a way that all <math>r_i</math> are distinct, and the number of <math>r_i</math> that are congruent to <math>3</math> modulo <math>4</math> is odd? These are precisely the <math>n</math> for which <math>A_n</math> is ''not'' ambivalent.
 We quickly see the following: