Proving a subgroup is equivalent to Z

In summary, the conversation discusses proving that the group (n^3+2n)Z + (n^4+3n^2+1)Z is equivalent to Z for n in N and n>=1. The discussion involves using the fact that subgroups of Z are of the form aZ and that aZ+bZ=dZ, as well as considering the Euclidean algorithm and polynomial division to find the gcd of the given polynomials. The conversation highlights the importance of confidence and trying different approaches to solving a problem.
  • #1
Gale
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2

Homework Statement



for [itex] n \in N, n \geq 1 [/itex] Prove that [itex] (n^{3} +2n)Z + (n^{4}+3n^{2}+1)Z= Z[/itex]

Homework Equations


I know subgroups of Z are of the form aZ for some a in Z and also that aZ+bZ= dZ, where d=gcd(a,b)

The Attempt at a Solution



So I was thinking if I could prove that the gcd of (n^3+2n) and (n^4+3n^2+1) was 1, then I could make the proof, but I'm struggling to figure out how to find a gcd of two polynomials... I also tried factoring to see if that led anywhere, but it didn't really...

Then I was thinking that if I could show that 1 was in the group, and since 1 generates Z, that would prove that the group was equivalent to Z... but then I wasn't actually sure that logic was sound.

Any help or some guidance in the right direction would be appreciated. Thanks!
 
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  • #2
Your first idea is correct. Do you know the Euclidean algorithm and polynomial division? So that you can check your work, you should find that

[tex] -(n^3+2n)(n^3+2n) + (1+n^2)(n^4+3n^2+1) = 1[/tex]
 
  • #3
Thank you! For some reason I didn't think polynomial long division was the way to go... I'm always so unconfident when I do these sorts of proofs.
 
  • #4
There is something to be said for confidence (or perhaps intuition?). Unfortunately, it seems like a lot of the time the only way to find that confidence/intuition is to fail/succeed at a tonne of questions.

Also, something you may notice is that (like this question) there are always many ways to attack a problem. We often cannot determine which way is correct until we have followed a path through to its conclusion and found that it is a dead end. Luckily, even dead-ends often provide insight and understanding into the structure of a problem: Even through failure, we are constantly learning.
 

FAQ: Proving a subgroup is equivalent to Z

What does it mean for a subgroup to be equivalent to Z?

When we say a subgroup is equivalent to Z, it means that it has the same structure and properties as the set of integers, also known as the group of integers under addition.

How do you prove that a subgroup is equivalent to Z?

To prove that a subgroup is equivalent to Z, we need to show that it satisfies the three defining properties of Z: closure, associativity, and existence of an identity element.

Can any subgroup be equivalent to Z?

No, not every subgroup can be equivalent to Z. For a subgroup to be equivalent to Z, it must satisfy the three defining properties of Z and have the same structure and properties as the set of integers.

What are the three defining properties of Z?

The three defining properties of Z are closure, associativity, and existence of an identity element. Closure means that the result of any operation on two elements in Z is also an element in Z. Associativity means that the grouping of elements in an operation does not affect the result. Existence of an identity element means that there is an element in Z that, when operated on any other element, results in that element unchanged.

Can we use any other method to prove that a subgroup is equivalent to Z?

Yes, there are other methods that can be used to prove that a subgroup is equivalent to Z, such as showing that it is isomorphic to Z or constructing a bijective homomorphism between the subgroup and Z.

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