By Randall R. Holmes

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**Example text**

Zm ⊕ Zn ∼ = Zmn if and only if gcd(m, n) = 1. Proof. (⇒) Assume that Zm ⊕ Zn ∼ = Zmn . Then there exists an isomorphism ϕ : Zmn → Zm ⊕ Zn . We have ϕ(1) = (a, b) for some a ∈ Zm and b ∈ Zn . Since the element 1 of Zmn has order mn, the element (a, b) of Zm ⊕ Zn must also have order mn. Let k = gcd(m, n). Then m = km and n = kn for some integers m and n . We have km n (a, b) = (km n a, km n b) = (n ma, m nb) = (0, 0). Since (a, b) has order mn and km n > 0 it follows that mn ≤ km n . So mn ≤ km n ≤ km kn = mn.

Ks ). Put I = {i1 , i2 , . . , ir } and K = {k1 , k2 , . . , ks }. To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle). Then σ and τ both fix m giving (στ )(m) = σ(τ (m)) = σ(m) = m = τ (m) = τ (σ(m)) = (τ σ)(m). Now assume that m ∈ I. Then σ(m) ∈ I, as well. Since I and K are disjoint by assumption, m and σ(m) are not in K, so they are fixed by τ .

The following theorem says that disjoint cycles commute. 1 Theorem. If σ and τ are disjoint cycles in Sn , then στ = τ σ. Proof. Let σ and τ be disjoint cycles in Sn . We can write σ = (i1 , i2 , . . , ir ) and τ = (k1 , k2 , . . , ks ). Put I = {i1 , i2 , . . , ir } and K = {k1 , k2 , . . , ks }. To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle).

### Abstract Algebra I by Randall R. Holmes

by Steven

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