We considered last time we can talk neatly about the conjugacy classes of the symmetric group and name them as the elements with the same cycle type. We have a special subset of the symmetric group called the alternating group.

**Defintion (Alternating Group).** The group of even permutations.

That seems a neat definition, but what does it mean for something to be an even or an odd permutation,

When we speak of numbers, I can make it very clear what I mean by even and odd. If I present to you \(3\), then you will instantly know it's odd. We can do a similar thing with permutations. We want to split them into a number and then we can determine their parity. Here is a claim,

**Claim 1.** Every permutation can be split into a series of disjoint transpositions.

This claim can be proven from thinking about a permutation and what a transposition is. I use transposition to mean swapping two elements. So \(\begin{pmatrix} 1 & 3 \end{pmatrix}\) is a transposition as all I'm doing is swapping \(1\) and \(3\). Let us quickly sketch a proof for this claim,

**Proof.** Take a permutation \(\sigma = \begin{pmatrix} a_1 & a_2 & \dots & a_n \end{pmatrix}\), we know from the last post that we can split this into a collection of disjoint cycles, say \(\sigma = \beta_1\beta_2 \dots \beta_k\) where these cycles \(\beta_i\) are just, \(\beta_i = \begin{pmatrix} b_1 & b_2 & \dots & b_\ell \end{pmatrix}\), but then this can be written as \(\beta_i = \begin{pmatrix} b_1 & b_2 \end{pmatrix} \begin{pmatrix} b_2 & b_3 \end{pmatrix} \dots \begin{pmatrix} b_{\ell-1} & b_\ell \end{pmatrix}\) and as all the \(\beta_i\)'s are disjoint and so are the transpositions we have a decomposition of disjoint transpositions. \(\blacksquare\)

It is rather easy to construct some sort of decomosition you just follow what we did for the \(\beta_i\)'s in the proof. For example \(\begin{pmatrix} 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}\) or another would be, \(\begin{pmatrix} 6 & 2 & 7 & 3 \end{pmatrix} = \begin{pmatrix} 6 & 2 \end{pmatrix} \begin{pmatrix} 2 & 7 \end{pmatrix}\begin{pmatrix} 7 & 3 \end{pmatrix}\) and we can notice very quickly that if the length of the permutation is odd, the partity is even and if the permutation is even, then the partity is odd. If we have a product of disjoint cycles we can notice that what will happen is we then have two lots of decompositions and that whole product is even if the total number of transpositions is even. In notation, if we have,

then we can say that this is even if \(n_1 + n_2 + \dots + n_q\) is odd, ie. sum of the number of elements in each cycle.

Now we define the alternating group,

**Definition 2 (Alternating Group).** The alternating group, denoted \(A_n\), is the group of even permutations.

Firstly, we can prove something quite nice about \(A_n\). We can prove that it is generated by the three cycles,

**Lemma 3.** \(A_n\) is generated by the \(3\)-cycles.

*Proof.* We firstly note that the \(3\)-cycles are in \(A_n\). Now let us take an element \(\sigma \in A_n\), and we write it in terms of it's transpostition decomposition,

and now we can consider every pair of transpositions, \(t_{2k-1}t_{2k}\) and see the following three cases,

- \(\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} a & b \end{pmatrix} = e\)
- \(\begin{pmatrix} a & c \end{pmatrix}\begin{pmatrix} a & b \end{pmatrix} = \begin{pmatrix} a & b & c \end{pmatrix}\)
- \(\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} c & d \end{pmatrix} = \begin{pmatrix} a & d & c \end{pmatrix}\begin{pmatrix} a & b & c \end{pmatrix}\)

The last one is slightly trixy, but we can see that the RHS = LHS and so we can rewrite two transpositions as some three cycles. Hence, we can multiply together the three cycles and get every different combination possible of transpostions and so \(A_n\) is generated by \(3\)-cycles.\(\blacksquare\)

Before we state and prove the final theorem, a few definitions for the road,

**Definition 4 (Transitive).** A action of a group on a set is called transitive when the set is nonempty and there is exactly one orbit.

So for \(S_n\) we can say that conjugation is transitive as we can pick some \(tau\) such that when we conjugate some \(\sigma\) that we map \(1\) to every other number in the set \(\{1, 2, \dots, n\}\).

Now we are ready to state and prove the splitting theorem,

**Theorem 5 (The Splitting Theorem).** The following are equivalent for some \(\sigma \in A_n\),

- the \(S_n\) conjugacy class of \(\sigma\) splits into two \(A_n\) conjugacy classes;
- there is no odd permutation that commutes with \(\sigma\);
- \(\sigma\) has no cycles of even lengths and they all have distinct lengths.

*Proof.* We shall aim to prove two things, firstly that the first and the second statement are equivalent and then the second and third are equivalent. Then by transitivity of implications we have that they all are equivalent. To prove the first two are equivalent, we argue as follows; as conjugation is transitive in \(S_n\), the stabiliser of some \(\sigma \in S_n\) is the centraliser of that element (all of the elements that commute with it). Let \(C_{S_n}(g)\) and \(C_{A_n}(g)\) be the cetraliser of \(g\) in \(S_n\) and \(A_n\) respectively, then \(C_{A_n}(g) = C_{S_n}(g) \cap A_n\) and so we can see that if the second condition holds then \(C_{A_n} = C_{S_n}\) and then \(|C_{A_n}| = \frac{|C_{S_n}|}{2}\) otherwise. Moreover, we can see that the conjugacy classes of \(S_n\) and \(A_n\) are of size, \(\frac{S_n}{C_{S_n}(g)}\) and \(\frac{S_n}{C_{A_n}(g)}\) which then yields that the first and second condition are equivalent.

Now we seek to prove that the last two statements are equivalent. We consider the cases of permuting two elements. If we consider that \(\sigma\) has a cycle of even length, then this cycle must be an odd permutation commuting with \(\sigma\), as \(\sigma \in A_n\). Now if we have two even permutations of equal length, say \(\ell\), then we can interchange these with a permutation that we can write as \(\ell\) transpositions, which must commute with \(g\). If we don't have either of these situations then we must have a permutation \(\tau\) that commutes with \(\sigma\) that fixes the cycles of \(\sigma\) and so \(\tau = \sigma^m, m \in \mathbb{Z}\) and so we can say that \(\tau\) is even. Hence, the last two are equivalent.\(\blacksquare\)

and so we can now tell when the conjugacy classes of \(A_n\) split into two, we also say that if they don't split then they are the same as \(S_n\), as \(A_n\) is of index 2. Let us take a simple example just to note what happens, consider \(A_3\). We can write \(A_3 = \{e, \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 3 & 2 \end{pmatrix}\}\) and we know that the conjugacy classes for \(S_n\) with respect to those elements is,

So now all the remains is to consider whether they split, so by the splitting theorem, we can look at the third condition and look at the second conjugacy class. We note that we have one odd cycle. Therefore, the second conjugacy class splits and we can write the conjugacy classes of \(A_3\) as,