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Compute all the conjugacy classes of $Alt(4)$

The order of $Alt(4)$ is 12, so do I really have to compute 144 conjugations and see what the classes are?. I was reading online and we can use the tetrahedron to fulfill this task. Does anyone know how to use it?

daniel
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  • See the following: http://math.stackexchange.com/questions/404656/splitting-of-conjugacy-class-in-alternating-group – Ege Erdil Aug 22 '16 at 00:19
  • You compute all the conjugacy classes of a group by conjugating every element with every element. It's going to be a tedious calculations if done by using brute force. – daniel Aug 22 '16 at 00:38
  • Do you know about the conjugacy classes in $S_n$? – anon Aug 22 '16 at 01:08
  • I have been reading very weird theorems that I have been never seen before. For example if two elements are conjugate in $A_n$ then they must be conjugate in $S_n$. There is one I find very useful which is two elements are conjugate if they have the same cycle type and there is other stuff about splitting a class because of some other stuff. – daniel Aug 22 '16 at 01:12
  • @daniel It is true that in the full symmetric group $S_n$, two elements are conjugate if and only if they have the same cycle structure. Since $A_n$ is a subgroup of $S_n$, two elements must have the same cycle type if they are conjugate, but the converse need not be true: two elements of the same cycle type are not necessarily conjugate. –  Aug 22 '16 at 01:47
  • @Bungo The following was the recommendation of my teacher: Burnside's orbit equation gives you a tool to connect the number of elements in the conjugacy class of the element x to the centralizer of x in G. Now you can work on both ends, creating elements in the conjugacy class, and limiting its cardinality by creating elements in the centralizer. – daniel Aug 22 '16 at 02:08

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First, note that in $S_4$ (or any $S_n$), two elements are conjugate if and only if they have the same cycle structure.

Since $A_4$ is a subgroup of $S_4$, two elements which are conjugate in $A_4$ are also conjugate in $S_4$, so they have the same cycle structure. However, the converse is not true: two elements with the same cycle structure are conjugate in $S_4$, but they need not be conjugate in $A_4$.

Start by listing the elements of $A_4$. We have $$1$$ $$(123), (132), (124), (142), (134), (143), (234), (243)$$ $$(12)(34), (13)(24), (14)(23)$$ The size of the conjugacy class containing an element $\alpha$ is equal to $|A_4 : G_{\alpha}|$, which is the index of the stabilizer (centralizer) of $\alpha$: $$G_{\alpha} = \{g \in A_4 : g\alpha g^{-1} = \alpha\}$$ Note that $|A_4 : G_{\alpha}| = |A_4|/|G_{\alpha}| = 12/|G_{\alpha}|$ so the size of each conjugacy class must be a divisor of $12$.

Note that $A_4$ has subgroups of index $1$, $3$, $4$, $6$, and $12$, so these are the only possible sizes of conjugacy classes.

Obviously $1$ is in its own conjugacy class.

There are $3$ elements of cycle type $(12)(34)$. Therefore the size of the conjugacy class containing $(12)(34)$ must be either $1$ or $3$. Note that if we conjugate $\alpha = (12)(34)$ using $g = (123)$ we get $g(12)(34)g^{-1} = (123)(12)(34)(132) = (14)(23)$, hence the size of the conjugacy class containing $\alpha$ cannot be $1$, and so it must be $3$. This means that the elements of type $(12)(34)$ are in a single conjugacy class, $\{(12)(34), (13)(24), (14)(23)\}$.

There are $8$ elements of cycle type $(123)$. Since $8$ is not a divisor of $12$, the size of the conjugacy class containing $(123)$ cannot be $8$, so there must be at least two conjugacy classes for this cycle type.

Also, all conjugacy classes for a given cycle type must have the same size. This is because if $\alpha$ and $\beta$ have the same cycle type, then they are conjugate in $S_4$, say $\beta = g\alpha g^{-1}$, and so if $G_{\alpha}$ is the stabilizer of $\alpha$ in $A_4$, then $g^{-1}G_{\alpha}g$ is the stabilizer of $\beta$, and it has the same order as $G_{\alpha}$. (Note that $g^{-1}G_{\alpha}g$ is a subgroup of $A_4$ even if $g \not\in A_4$, since $G_{\alpha}$ is a subgroup of $A_4$, and $A_4$ is normal in $S_4$.)

There are $8$ elements of cycle type $(123)$, so the conjugacy class containing $\alpha = (123)$ must have size $1$, $3$, $4$, or $6$. But of these, only $1$ and $4$ are divisors of $8$, so they are the only possibilities, since the $8$ elements of this cycle type must be partitioned into conjugacy classes of equal size.

If we conjugate $\alpha$ by $g = (124)$, we get $g(123)g^{-1} = (124)(123)(142) = (243)$, so the size of the conjugacy class containing $(123)$ is not $1$, therefore it is $4$.

How do we determine which elements are conjugate to $(123)?$ Any such element must be of the form $g(123)g^{-1}$, where $g \in A_4$ and therefore $g$ can be written as the product of an even number of transpositions. Conjugation by a transposition swaps two digits in the full cycle representation $(123)(4)$, so conjugation by an even number of transpositions causes an even number of swaps.

Since $(142)(3)$, $(134)(2)$, and $(243)(1)$ can each be obtained from $(123)(4)$ by performing two swaps, the conjugacy class containing $(123)$ is $\{(123), (142), (134), (243)\}$.

Similarly, the conjugacy class containing $(132)$ is $\{(132), (124), (234), (143)\}$.

Note that we could have shortened the argument considerably by using this result about splitting of conjugacy classes in the alternating group. This tells us that each conjugacy class in $S_4$ is either a single conjugacy class in $A_4$, or it is split evenly into two classes. We can use this argument as follows.

In $S_4$, two elements are conjugate if and only if they have the same cycle type. So the class containing $(12)(34)$ has size $3$. Since this can't be split evenly into two classes, the same conjugacy class must exist in $A_4$.

In $S_4$, the class containing $(123)$ has size $8$. Since this is impossible in $A_4$ as $8$ is not a divisor of $12$, this class must be split in half by $A_4$ to form two classes of size $4$.

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  • Thank you so much for such elaborated answer. Now I am gonna take a while to understand every single idea. The first thing to understand is how we view orbits and stabilizers. Based on my book the orbit $O_x={gx | g \in G}$ and $stab_G(x)={g\in G |gx=x}$ , so basically $g*x$ is the operation or the function acting on the set right? – daniel Aug 22 '16 at 03:18
  • @daniel Note that here we are considering the action by conjugation, so keep in mind that $g*x = gxg^{-1}$. And the stabilizer of $x$ under this action is the set of all $g$ such that $gxg^{-1} = x$, or equivalently $gx = xg$, i.e. the set of all $g$ which commute with $x$, which is simply the centralizer of $x$. –  Aug 22 '16 at 03:22
  • "there must be either one or three conjugacy classes of this cycle type." Can you explain this – daniel Aug 22 '16 at 03:57
  • @daniel Actually that's a true statement, but it's not what I meant to say. I'll edit to fix now, so hopefully it will make more sense. –  Aug 22 '16 at 04:43