Conjugacy Classes of A4

  • #1
130
1

Homework Statement



Determine the conjugacy classes for A4, the set of even permutations of S4.


The Attempt at a Solution



I'm trying to figure out a correct way that doesn't involve much straight up computation.

Here is my thinking;

Elements being conjugate in A4 mean they are conjugate in S4. From a Lemma in my notes only elements of the same cyclic type can be conjugate. Meaning conjugacy in A4 can only happen between elements of the same cyclic type.

A4 = {(1), (123), (124), (132), (134), (142), (143), (234), (243), (12)(34), (13)(24), (14)(23)}

So in S4, {(12)(34), (13)(24), (14)(23)} is a conjugacy class of order 3.

Now in A4 either this conjugacy class remains, or it splits in to two conjugacy classes of order 2 and 1, or three classes of order 1. To be of order one means that element is in the centre of A4, but this is easily checked not to be true for (12)(34), (13)(24), and (14)(23); so {(12)(34), (13)(24), (14)(23)} is a class in A4.


The last part remains. In S4 the centraliser of (123) is {(1), (123) and (132)} and hence this is also the centraliser in A4. Since the order of the conjugacy class is equal to the index of the centraliser of that element the class of (123) must have 4 elements. Also the centraliser of (132) is the same set and so the class of (132) must also have 4 elements.

Now direct computation shows that the class of (123) is {(123), (243), (142), (134)}

As (132) is not in this, the class of (132) must consists of the 4 remaining cycles of length 3.


So the classes are

{(1)}
{(123), (243), (142), (134)}
{(132), (124), (143), (234)}
{(12)(34), (13)(24), (14)(23)}



This proof works but my problem lies in having to compute the class for (123). It required me to try calculating the conjugates of (123) using random elements of A4 which takes a while and doesn't always produce unique results. Is there a faster way that I'm missing?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
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Homework Statement



Determine the conjugacy classes for A4, the set of even permutations of S4.


The Attempt at a Solution



I'm trying to figure out a correct way that doesn't involve much straight up computation.

Here is my thinking;

Elements being conjugate in A4 mean they are conjugate in S4. From a Lemma in my notes only elements of the same cyclic type can be conjugate. Meaning conjugacy in A4 can only happen between elements of the same cyclic type.

A4 = {(1), (123), (124), (132), (134), (142), (143), (234), (243), (12)(34), (13)(24), (14)(23)}

So in S4, {(12)(34), (13)(24), (14)(23)} is a conjugacy class of order 3.

Now in A4 either this conjugacy class remains, or it splits in to two conjugacy classes of order 2 and 1, or three classes of order 1. To be of order one means that element is in the centre of A4, but this is easily checked not to be true for (12)(34), (13)(24), and (14)(23); so {(12)(34), (13)(24), (14)(23)} is a class in A4.


The last part remains. In S4 the centraliser of (123) is {(1), (123) and (132)} and hence this is also the centraliser in A4. Since the order of the conjugacy class is equal to the index of the centraliser of that element the class of (123) must have 4 elements. Also the centraliser of (132) is the same set and so the class of (132) must also have 4 elements.

Now direct computation shows that the class of (123) is {(123), (243), (142), (134)}

As (132) is not in this, the class of (132) must consists of the 4 remaining cycles of length 3.


So the classes are

{(1)}
{(123), (243), (142), (134)}
{(132), (124), (143), (234)}
{(12)(34), (13)(24), (14)(23)}



This proof works but my problem lies in having to compute the class for (123). It required me to try calculating the conjugates of (123) using random elements of A4 which takes a while and doesn't always produce unique results. Is there a faster way that I'm missing?

Random elements of A4 certainly isn't the best way. There are patterns here. To go from (123) to (243) requires you change 1 into 2 and 2 into 4. So clearly (12)(24)(123)(24)(12)=(243). It involves two substitutions. And (12)(24) is in A4. You can get the others in the class the same pattern. Now going from (123) to (132) takes one substitution, 2 into 3. So (23)(123)(23)=(132). But (23) is not in A4. Think "number of transpositions needed".
 

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