Conjugacy classes of the Alternating Group A_{n}

GCS: 17-xi-03

This note is to correct some insanity in Friday’s lecture. In what follows, n is a positive integer and at least 2.

First we recall a general fact which we proved via working on a problem sheet: Lemma Suppose that G is a group and that H and K are subgroups of G, then |K : H ∩ K| ≤ |G : H|.

Suppose that x ∈ A_{n} ≤ S_{n}. Let the conjugacy class of x in S_{n} be C and the conjugacy class of x in A_{n} be D. Theorem C = D if and only x commutes with an odd element of S_{n}. If this is not the case, then 2|D| = |C|. Proof Apply the lemma with S_{n} = G, A_{n} = H and C_{S_{n}}(x) = K. Note that C_{A_{n}}(x) = A_{n} ∩ C_{S_{n}}(x) so |C_{S_{n}}(x) : C_{A_{n}}(x)| ≤ |S_{n} : A_{n}| = 2.

Now if this index is 1 if and only if C_{S_{n}}(x) = C_{A_{n}}(x) which happens if and only if x commutes with no odd element of S_{n}. In these circumstances |S_{n} : A_{n}| · |A_{n} : C_{A_{n}}(x)| = |S_{n} : C_{A_{n}}(x)| = |S_{n} : C_{S_{n}}(x)| and so 2|D| = |C|.

On the other hand if |C_{S_{n}}(x) : C_{A_{n}}(x)| = 2, then a factor of two can be cancelled from |S_{n} : A_{n}| · |A_{n} : C_{A_{n}}(x)| = |S_{n} : C_{A_{n}}(x)| = |S_{n} : C_{S_{n}}(x)| · |C_{S_{n}}(x) : C_{A_{n}}| to yield that |C| = |D|, but D ⊆ C so C = D. Conversely if C = D, then |S_{n} : C_{S_{n}}(x)| = |A_{n} : C_{A_{n}}(x)|. Now |S_{n} : A_{n}| · |A_{n} : C_{A_{n}}(x)| = |S_{n} : C_{A_{n}}(x)| = |S_{n} : C_{S_{n}}(x)| · |C_{S_{n}}(x) : C_{A_{n}}| so |S_{n} : A_{n}| = |C_{S_{n}}(x) : C_{A_{n}}| = 2.

The proof is complete. Example In S_{3} the elements (1, 2, 3) and (1, 3, 2) are conjugate. In A_{3} = ⟨(1, 2, 3)⟩ they are not (the group A_{3} is abelian so different elements are never conjugate). The conjugacy class of (1, 2, 3) is S_{3} is {(1, 2, 3), (1, 3, 2)} whereas is A_{3} it is {(1, 2, 3)}. Notice that the centralizer of (1, 2, 3) in S_{3} is ⟨(1, 2, 3)⟩ so (1, 2, 3) commutes with no odd elements of S_{3}. Observations In the lecture I speculated that when a conjugacy class of S_{n} falls into two conjugacy classes in A_{n}, then the elements of one class might necessarily be the inverses of the elements of the other. Note that inverse elements do have the same cycle shape since you can invert an element in standard form by reversing each of its cycles. This speculation turns out to be false. In A_{5} there are two conjugacy classes of elements with shape which is a 5-cycle. However g = (2, 5)(3, 4) ∈ A_{5} and g^{−1}(1, 2, 3, 4, 5)g = (1, 5, 4, 3, 2) = (1, 2, 3, 4, 5)^{−1}.