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A permutation τ ∈ Sn which interchanges two elements of n and leaves the rest fixed is called a transposition. 11. Let σ ∈ Sn . Then there are transpositions τ1 , . . , τk such that σ = τ1 · · · τk . 1) (i1 i2 . . ir ) = (i1 ir ) · · · (i1 i3 )(i1 i2 ). 12. Decompose σ= 1 2 3 4 5 2 5 3 1 4 ∈ S5 into a product of transpositions. Solution. We have σ = (3)(1 2 5 4) = (1 2 5 4) = (1 4)(1 5)(1 2). Some alternative decompositions are σ = (2 1)(2 4)(2 5) = (5 2)(5 1)(5 4). 5. Symmetry groups Let S be a set of points in Rn , where n = 1, 2, 3, .

The following sets are countably infinite. a) b) c) d) Any infinite subset S ⊆ N0 . X ∪ Y where X, Y are countably infinite. X ∪ Y where X is countably infinite and Y is finite. The set of all ordered pairs of natural numbers N0 × N0 = {(m, n) : m, n ∈ N0 }. 56 4. FINITE AND INFINITE SETS, CARDINALITY AND COUNTABILITY e) The set of all positive rational numbers Q+ = a : a, b ∈ N0 , a, b > 0 . b Solution. a) Since S is infinite it cannot be empty. Let S0 = S. By WOP, S0 has a least element s0 say.

N}. 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!

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