By Irena Swanson

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Additional resources for Abstract Algebra [Lecture notes]

Sample text

Proof. Let ϕ : Un1 ···nk → Un1 ⊕ Un2 ⊕ · · · ⊕ Unk be the restriction of ϕ above. We first need to verify that the map is well-defined, namely that the image of ϕ is in Un1 ⊕ Un2 ⊕ · · · ⊕ Unk . Indeed, if m is relatively prime to n1 · · · nk , it is relatively prime to each ni . Thus ϕ is well-defined. Since this map is a restriction of an injective map, it is injective. It is easy to verify that it is a group homomorphism (this does not follow from ϕ being a homomorphism, because the two operations are different), and to prove surjectivity, let (b1 , .

But it couldn’t be anything smaller, for otherwise some gil wouldn’t be e. 5 Let G and H be finite groups such that G ⊕ H is cyclic. Then G and H are cyclic and |G| and |H| are relatively prime. Proof. Suppose that G ⊕ H is generated by (a, b). Then G is generated by a, so G is cyclic. Similarly H = b is cyclic. By the previous theorem, |a| · |b| = |G| · |H| = |G ⊕ H| = |(a, b)| = lcm{|a|, |b|}, so that the orders of a and b are relatively prime. Thus the orders of G and H are relatively prime.

Let ϕ : Un1 ···nk → Un1 ⊕ Un2 ⊕ · · · ⊕ Unk be the restriction of ϕ above. We first need to verify that the map is well-defined, namely that the image of ϕ is in Un1 ⊕ Un2 ⊕ · · · ⊕ Unk . Indeed, if m is relatively prime to n1 · · · nk , it is relatively prime to each ni . Thus ϕ is well-defined. Since this map is a restriction of an injective map, it is injective. It is easy to verify that it is a group homomorphism (this does not follow from ϕ being a homomorphism, because the two operations are different), and to prove surjectivity, let (b1 , .