Group Theory: Showing the Order of Element 2 is 2k

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For this, you can use the fact that p is a prime number and the elements 2^1, 2^2, ... , 2^k are all distinct modulo p. This means that the order of 2 must divide p-1, which gives you the first part of the proof.For the second part, you are to deduce that k=2^n for some natural number n. To do this, you can use the fact that the order of 2 is 2k and then use the properties of exponents to show that k must be a power of 2.
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Homework Statement



Let p=2^(k)+1 , in which k is a positive integer, be a prime number. Let G be the group of integers 1, 2, ... , p-1 under multiplication defined modulo p.

By first considering the elements 2^1, 2^2, ... , 2^k and then the elements 2^(k+1), 2^(k+2), ... show that the order of the element 2 is 2k.

Deduce that k= 2^n for n is a natural number.

Homework Equations



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The Attempt at a Solution



My problem is that I don't understand the question, or how to get started. I'm looking for some hints on how to approach this. I figured that the remainder of (2^k)/(p^(k)+1) should be zero for k to be the order. Is this right? I really have no idea what I'm supposed to do ;/


This is my first time posting on here so I hope I've set this out right! Thanks in advance :)
 
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For the first part, you are to show that order of the element 2 is 2k, which means
Show -
[tex]2^{2k} (\tex{mod} p) \equiv 1[/tex]
 

FAQ: Group Theory: Showing the Order of Element 2 is 2k

What is Group Theory?

Group Theory is a branch of mathematics that studies the properties of groups, which are mathematical structures that consist of a set of elements and a binary operation that combines two elements to form a third element. Groups are used to describe symmetries and transformations in various areas of science, including physics, chemistry, and computer science.

What does it mean to show the order of an element is 2k?

The order of an element in a group is the smallest positive integer k such that the element multiplied by itself k times results in the identity element (the element that does not change any other element when combined with it). Showing the order of an element is 2k means demonstrating that the element multiplied by itself 2k times is equal to the identity element.

Why is it important to show the order of element 2 is 2k?

Showing the order of an element is important because it helps to understand the structure and properties of the group. It also allows us to classify groups and determine if they are isomorphic (having the same structure) to other groups. In particular, showing the order of element 2 is 2k is important because it helps to identify if the group is a cyclic group (a group generated by a single element) or not.

How is the order of element 2 related to the order of the group?

The order of the group, denoted as |G|, is the number of elements in the group. The order of element 2, denoted as |2|, is the smallest positive integer k such that 2^k = e (where e is the identity element). The order of element 2 is always a divisor of the order of the group, that is, |2| divides |G|. This relationship is known as Lagrange's theorem and is a fundamental result in group theory.

Can you give an example of showing the order of element 2 is 2k?

Yes, consider the group of integers modulo 6 under addition, denoted as (Z/6Z, +). The elements of this group are {0, 1, 2, 3, 4, 5}, and the identity element is 0. We can show that the order of element 2 is 2k by calculating 2^1 = 2, 2^2 = 4, 2^3 = 2 (mod 6), and 2^4 = 4 (mod 6), and so on. It turns out that 2^3 = 2 (mod 6) is the first time we get back to the identity element, so the order of element 2 is 3, which is equal to 2k = 2*3.

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