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# symmetric group

 Definition/Summary The symmetric group S(n) or Sym(n) is the group of all possible permutations of n symbols. It has order n!. It has an index-2 subgroup, the alternating group A(n) or Alt(n), the group of all possible even permutations of n symbols. That group has order n!/2. For n >= 5, the alternating group is simple.

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 Extended explanation First, some details about permutations. Every finite-length permutation can be broken up into cyclic permutations, where a cyclic permutation has the form a1 -> a2, a2 -> a3, a3 -> a4, ..., a(n) -> a1 where n is the length of the cycle. The conjugacy classes of S(n) are all elements that share some set of lengths of cycles. For S(3), the elements are 123 -> 123, 132, 213, 231, 312, 321. Its conjugacy classes are 13: {123} 1*2: {132, 213, 321} 3: {231, 312} The elements of A(n) are all even permutations, those with an even number of even-length cycles. Thus, A(3) has elements 123, 231, 312. Some of the S(n) conjugacy classes get split in two, those with all odd cycle lengths that do not get repeated. Thus, the 3 splits into {231} and {312}. The remaining coset of A(n) in S(n) is the odd permutations, those with an odd number of even-length cycles. The group (even permutations, odd permutations) is, of course, Z(2). The representation theory of the symmetric group is very elegant, with the irreducible representations or irreps being labeled with sets of cycle lengths, just like the classes. The character values are all integers. Going from a symmetric group to its alternating group, the irreps with length sets that are dual to each other get merged, while the irreps with self-dual length sets get split. Duality of the length sets is defined as follows: Order the lengths from largest to smallest, then make (length) boxes for each one in a 2D pattern. Find out how many boxes there are in the other direction, and that gives the dual. Example: The length set 3,2,1,1 gives *** ** * * Its dual is 4,2,1 For even permutations, duality leaves the character value's sign unchanged, while for odd permutations, duality reverses the sign. In the self-dual case, odd permutations get zero character. The irrep (n) for S(n) is the identity irrep, while the irrep (1n) is the parity irrep; it is 1 for even permutations and -1 for odd permutations. For S(3), the 13 and 3 irreps are a dual set, while the 1*2 irrep is self-dual. Its character table is $\begin{matrix} & 3 & 1 \cdot 2 & 1^3 \\ 1^3 & 1 & 2 & 1 \\ 1 \cdot 2 & 1 & 0 & -1 \\ 3 & 1 & -1 & 1 \end{matrix}$ Rows: classes, columns: irreps For A(3), we get the character table $\begin{matrix} & (3,1^3) & (1 \cdot 2)_1 & (1 \cdot 2)_2 \\ 1^3 & 1 & 1 & 1 \\ 3_1 & 1 & \frac12(-1 + \sqrt{-3}) & \frac12(-1 - \sqrt{-3}) \\ 3_2 & 1 & \frac12(-1 - \sqrt{-3}) & \frac12(-1 + \sqrt{-3}) \end{matrix}$ The sqrt(-3) is typical of splitting.