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The Sn = Perm(n) is called the symmetric group on n objects or the symmetric group of degree n or the permutation group on n objects. 5. Sn has order |Sn | = n!. Proof. Defining an element σ ∈ Sn is equivalent to specifying the list σ(1), σ(2), . . , σ(n) consisting of the n numbers 1, 2, . . , n taken in some order with no repetitions. To do this we have • n choices for σ(1), • n − 1 choices for σ(2) (taken from the remaining n − 1 elements), • and so on. In all, this gives n × (n − 1) × · · · × 2 × 1 = n!

StabG (xU )| If there is only one orbit, then the action is said to be transitive, and in this case, for any x ∈ X we have X = OrbG (x) and |X| = |G|/| StabG (x)|. Given an action of (G, ∗) on X, another useful idea is that of the fixed point set or fixed set of an element g ∈ G, FixG (g) = {x ∈ X : gx = x}. FixG (g) is also often denoted X g . 7. 37 (Burnside Formula). If (G, ∗) acts on X with G and X finite, then number of orbits = 1 |G| | FixG (g)|. g∈G Proof. 34) U = OrbG (x) an orbit |G| U an orbit 1 = U an orbit = number of orbits.

If g is infinite then g is said to have infinite order |g| = ∞. 20. If g ∈ G has finite order |g| then |g| = min{m ∈ Z+ : m > 0, g m = ι}. 21. In the group Sn the cyclic permutation (i1 i2 · · · ir ) of length r has order |(i1 i2 · · · ir )| = r. Solution. Setting σ = (i1 i2 · · · ir ), we have σ k (1) = hence |σ| r. As ik = 1 for 1 < k ik+1 if k < r, i1 if k = r, r, r is the smallest such power which is ι, hence |σ| = r. So for example, |(1 2)| = 2, |(1 2 3)| = 3 and |(1 2 3 4)| = 4. But notice that the product (1 2)(3 4 5) satisfies ((1 2)(3 4 5))2 = (1 2)(3 4 5)(1 2)(3 4 5) = (3 5 4), hence |(1 2)(3 4 5)| = 6.

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