MATH 351 – Modern Algebra I Selected Solutions for Problem Set 9 November 30, 2018

6.26. By Theorem 6.6 in Lee, we know that every σ ∈ S_{n} is a product of transpositions, but those transpositions can contain any of the (ab), with 1 ≤ a < b ≤ n. On the other hand, by Lemma 6.2, we know that

(1a)(1b)(1a) = (1a)(1b)(1a)^{−1} = (ab).

Hence in any factorization of σ into transpositions, we can replace any transposition (ab) with a, b ≠ 1 by the product (1a)(1b)(1a) as above. This shows that every σ ∈ S_{n} can be written as a product of the “special” transpositions (1i) with 2 ≤ i ≤ n.

6.28. There are many different proofs of this fact. Let’s look at two, both of which are based on Exercise 6.26 above.

Proof 1: (This one might be the “slickest”.) Letting σ = (12) and τ = (12 ⋯ n) as in the problem statement, note that Lemma 6.2 implies:

\begin{aligned}τστ^{−1} & = (23) \\ τ(23)τ^{−1} & = τ^{2}στ^{−2} = (34) \\ τ(34)τ^{−1} & = τ^{3}στ^{−3} = (45), \end{aligned}

and hence, continuing in the same way every “consecutive” transposition (k k + 1) with 1 ≤ k ≤ n − 1 can be written as required by the problem. Then, using Lemma 6.2 again, note that

\begin{aligned}(23)(12)(23) & = (13) = (τστ^{−1})σ(τστ^{−1}) \\ (34)(13)(34) & = (14) = (τ^{2}στ^{−2})(τστ^{−1})σ(τστ^{−1})(τ^{2}στ^{−2}) \end{aligned}

and similarly, all of the “special transpositions” (1i) from Exercise 6.26 can be written as products of powers of σ and τ. It follows from Exercise 6.26, then, that every ρ ∈ S_{n} can be written in this way too.

Proof 2a: A second way to derive the same conclusion (using Exercise 6.26 in a similar way), is to alternate conjugating 2-cycles and n-cycles, like the following: We have, by Lemma 6.2,

\begin{aligned}στσ^{−1} & = (2134 ⋯ n) = τ_{2} \\ τ_{2}στ_{2}^{−1} & = (31) = (13) \\ (13)τ_{2}(13) & = (2314 ⋯ n) = τ_{3} \\ τ_{3}(13)τ_{3}^{−1} & = (41) = (14) \end{aligned}

and continuing in the same way, we get all the (1i) with 2 ≤ i ≤ n as products of powers of σ and τ.

Proof 2b: Proof 2a can also be phrased as an argument by induction as follows. We want to show that in S_{n}, all of the “special” transpositions (1i) for 2 ≤ i ≤ n can be obtained as products of powers of σ and τ. The base case is i = 2 and there is nothing to prove since σ = (12). So now let’s assume that we have shown (12), (13), …, (1k) can be written this way. Consider (1 k + 1). Then, following what we did in Proof 2a, we consider the product giving the element we called τ_{k} above:

τ_{k} = (1k)(1 k − 1) ⋯ (13)(12)τ(12)(13) ⋯ (1 k − 1)(1k)

By Lemma 6.2, since the product on the right of τ is the inverse of the product on the left of τ, this equals

τ_{k} = (2 3 4 ⋯ k 1 k + 1 k + 2 ⋯ n).

(In the n-cycle τ, we’re transposing the 1 and the 2, then the 1 and the 3, etc. up to the 1 and the k, so 2, 3, …, k all end up to the left of 1, and the remaining numbers k + 1, …, n have not moved.) By Lemma 6.2 yet again, then

τ_{k}(1k)τ_{k}^{−1} = (k + 1 1) = (1 k + 1)

is a product of powers of τ and σ, since τ_{k} and (1k) are such products (that’s the induction hypothesis) and this finishes the proof.

7.8. We want to show that if a is odd order, then C(a) = C(a^{4}). We’ll set this up as a proof showing each of these sets is contained in the other.

⊆: Let g ∈ C(a). Then ga = ag, so

ga^{4} = (ga)a^{3} = (ag)a^{3} = a(ga)a^{2} = a(ag)a^{2} = a^{2}(ga)a = a^{3}ga = a^{4}g.

Hence g ∈ C(a^{4}), so C(a) ⊆ C(a^{4}). (Note: We don’t need the hypothesis that a has odd order for this inclusion.)

⊇: Now let g ∈ C(a^{4}). By definition this means ga^{4} = a^{4}g. We have to show this implies ga = ag when |a| is odd. Let k = |a|. Then by integer division we have k = 4q + r with 0 ≤ r < 4. But note r = 0, 2 are not possible, since then k would be even. This says there are two possible values for r, namely r = 1 and r = 3. We handle each of those cases separately.

If r = 1, then a^{4q+1} = e, so a = a^{−4q}. Note that a^{−4q} = (a^{4})^{−q}. Since we have ga^{4} = a^{4}g, we also have a^{−4}g = ga^{−4} by multiplying both sides of the previous equation by a^{−4} on the left and the right. It follows that ga^{−4q} = a^{−4q}g for all integers q. Hence

ga = ga^{−4q} = a^{−4q}g = ag

and g ∈ C(a). This shows C(a^{4}) ⊆ C(a) when |a| = 4q + 1.

If r = 3, then a^{4q+3} = e, so a = a^{4(q+1)}. Note that a^{4(q+1)} = (a^{4})^{q+1}. If g ∈ C(a^{4}), then it follows that ga^{4(q+1)} = a^{4(q+1)}g. Hence

ga = ga^{4(q+1)} = a^{4(q+1)}g = ag

and g ∈ C(a). This shows C(a^{4}) ⊆ C(a) when |a| = 4q + 3 as well.