Christian AlexanderMath 403, Professor Haines

1) The elements of S3 are the identity e, the three transpositions τ , and the twopermutations σ of all three elements. We know from class that τ, σ 6= e, andthat τ2 = σ3 = e. Thus τ3 = τ 6= e; if σ2 were equal to e, then we’d haveσ = σ3 ◦ σ = σ2 ◦ σ2 = e, which is false. Thus, e and the three τ are the fourα ∈ S3 with α2 = e, and e and the two σ are the three α with α3 = e.

2) It is easy to see that any α ∈ Sn with α2 = e, α 6= e must be a product of disjointtranspositions. (Proof: Choose any a ∈ Sn. Then α(a) = b, and α(b) = a, forsome b. For at least one a ∈ Sn, we will have a 6= b, since otherwise α = e. Thus,we can write α as the product of all the non-trivial transpositions (a b) taken overa ∈ Sn, which are of course disjoint.) If α4 = e, α 6= e, then either α2 = e or not;if so, then the above characterizes α. If not, then it characterizes β = α2; i.e., βis a product of disjoint transpositions. It’s straightforward but tedious to showthat α2 = (a b) is impossible; thus, β ∈ S4 must be of the form (a b)(c d) (if anyexist). Since β = α2, we can read off that α(a) = x, α(x) = b, α(b) = y, α(y) = a;we must have a, b, x, y all distinct, so this says that α is a 4-cycle. Of course, ifα ∈ S4 is a 4-cycle, then α2 is the product of two disjoint transpositions; so theset of all α such that α4 = e, α2 6= e is the set of all 4-cycles. Finally, we maycharacterize all α such that α4 = e: they are e, all products of disjoint 2-cycles,and all 4-cycles.

3) (i) is trivially false: there are n! permutations of n elements, and each elementof Sn is just one such permutation. (ii) is true; every finite group necessarilyhas this property. (iii) is true; this is the product on Sn. (iv) is false; (1 2) ◦(2 3) 6= (2 3) ◦ (1 2). (v) is false; (1 2) and (2 3) are both 2-cycles, whereas(1 2) ◦ (2 3) is a 3-cycle. (vi) is true; if γ = (i1 · · · ir) is an r-cycle, then αγα−1

is also an r-cycle, since αγα−1 sends α(ij) → α(ik) if γ sends ij → ik, and soαγα−1 = (α(i1) · · · α(ir)) explicitly. (vii) is the opposite of true; transpositionsare odd, since (−1)1 is negative. (viii) is true; this is a particular case of theparity of permutations. (ix) is false; we can write (1 2 3 4) = (1 2)(1 3)(1 4) =(1 2)(1 3)(2 4)(1 4)(1 2). (x) is trivially false; setting α = e, we can put σ andω as any two distinct transpositions.

4) For part (a): the number of ways to choose r numbers from n is the product(n) · · · (n− r + 1). Each choice determines an r-cycle up to cyclic permutationof the elements; since r cyclic permutations are possible, the number of cyclesis 1

r (n) · · · (n− r + 1). For part (b): in an entirely similar way to part (a), thenumber of ways of creating k r-cycles will be 1

rk(n) · · · (n − kr + 1), since this

will be the number of ways to pick kr elements, i.e. k groups of r elements, fromwhich we obtain 1

rk(n) · · · (n − kr + 1) products of unique r-cycles. The only

worry is overcounting based on re-ordering of the cycles, which is accounted forby the k! in the denominator.

5) For part (a): letting ik be any element in the cycle, αr(ik) = αk(αr−k(ik)) =

1

αk(i1) = ik. For part (b), observe that q < r says that ir−q 6= i1, and αq(ir−q) =i1, so that αq 6= e.

6) By induction: the base cases k = 1 and k = 2 are obvious, and if the claimholds for k, then αk+2βk+2 = αk+1βαβk+1 = · · · = αβαkβkαβ = αβ(αβ)kαβ =(αβ)k+2. Therefore the claim holds for k generally.

7) Suppose α isn’t the identity; then there must be some a 6= b such that α(a) = b.Let c differ from both a and b; then commutativity implies that (a c)α(a c) = α;but passing c into each side, we come out with b on the LHS, and α(c) on theother. If α(c) 6= b, we have a contradiction; if they are equal, then α(a) = α(c)says a = c, also a contradiction.

8) The system of (a) fails to be a group because the operation is not associative,since (a − b) − c 6= a − (b − c) in general for integers. The system of (b) alsofails, since almost no elements have an inverse under the given operation. Thesystem of (c) works fine as a group; it’s isomorphic to (Z7,+). The system of(d) works fine, since it’s a closed subset, thus subgroup, of the ordinary groupof rational numbers.

9) A polynomial in Z[x] corresponds uniquely to a tuple (n, a0, a1, ...an) of a pos-itive integer n and integers a0, a1, ...an. Now consider the number of tupleswhose coefficients satisfy n + |a0| + |a1| + ... + |an| = N for some fixed N (letT (N) denote the set of all such tuples). Finding the number of solutions ton+ b0 + b1 + ...+ bn = N with bi ≥ 0 for 0 ≤ i ≤ n is equivalent to finding thenumber of solutions to (b0 + 1) + (b1 + 1) + ...+ (bn + 1) = N ; and this is

(N−1n

)(we are partitioning N by n+ 1 positive integers, so a stars-and-bars argumentapplies). Now, there are at most 2n+1 different tuples (a0, a1, ...an) of integers,not necessarily non-negative, corresponding to any partition (|a0|, |a1|, ...|an|);so we find that |T (N)| ≤

∑Nn=1 2n+1

(N−1n

). In particular, T (N) is finite. Now,

any tuple obviously belongs to T (N) for some N ; so we find that the number ofpolynomials in Z[x] is in correspondence with (a subset of)

⋃N∈N T (N). This

set is a countable union of finite sets, and so is countable; and therefore theinteger-coefficient polynomials must be countable as well.

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