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The conjugacy classes of $A_{5}$ : sizes $1, 20, 15, 12, 12$ and the split $5$ -cycles

Example

The five conjugacy classes of $A_{5}$ have representatives and sizes

1 : 1, (123) : 20, (12) (34) : 15,

(12345) : 12, (13524) : 12.

The last two classes are the two halves of the $S_{5}$ class of $5$ -cycles.

Facts & Assumptions

Given: The alternating group $A_{5}$ .

[F1]

$S_{n}$ -classes are indexed by the tuples with $\sum k c_{k} = n$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ ); a permutation of type $(c_{k})$ has centralizer cardinality $\prod_{k} k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ); and $n! = \sum n! / \prod_{k} k^{c_{k}} c_{k}!$ over those tuples, the summand indexed by $(c_{k})$ being the size of the corresponding class (The class equation of $S_{n}$ is $n! = \sum_{\sum k c_{k} = n} n! / \prod_{k} k^{c_{k}} c_{k}!$ ).

[F2]

A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are included as $1$ -cycles (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[F3]

For $n \geq 2$ and $σ \in A_{n}$ , the $S_{n}$ -class of $σ$ splits into two $A_{n}$ -classes of equal size exactly when all cycle lengths in its decomposition, including $1$ -cycles for fixed points, are odd and no two are equal (For $n \geq 2$ , an $S_{n}$ -class of an even permutation splits in $A_{n}$ exactly when all cycle lengths, including $1$ -cycles, are odd and distinct).

[F4]

$∣ A_{5} ∣ = 5! / 2 = 60$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

Verification

technique · counting

1.1

By [F1] and [F2], the even $S_{5}$ types are $1^{5}$ , $3, 1^{2}$ , $2^{2}, 1$ , and $5$ , with symmetric class sizes $1, 20, 15, 24$ .

F1F2algebra

2.1

By [F3], the first three stay single classes, while the $5$ -cycle class splits into two equal classes of size $12$ .

F3step 1.1

3.1

Put $s = (12345)$ . The permutation $q = (2 3 5 4)$ satisfies $q s q^{- 1} = s^{2} = (13524)$ and is odd by [F2]. Every other conjugator from $s$ to $s^{2}$ differs from $q$ by an element centralizing $s$ ; such a centralizer element is determined by the image of $1$ and is therefore a power of the even $5$ -cycle $s$ . Thus every conjugator is odd, so $s$ and $s^{2}$ lie in the two different halves from step 2.1.

F2step 2.1algebra

4.1

The total $1 + 20 + 15 + 12 + 12 = 60$ agrees with [F4].

F4step 1.1step 2.1∎

Depends on

Used by

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Sources

K. Conrad, Conjugacy Classes (standard reference, not scraped)

The conjugacy classes of A5A5​: sizes 1,20,15,12,121,20,15,12,12 and the split 55-cycles