Understanding Number Theory Proofs: Order of Elements in Finite Groups

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Discussion Overview

The discussion revolves around understanding proofs related to number theory, specifically focusing on the order of elements in finite groups and the implications of Euler's and Fermat's theorems. Participants are exploring the relationships between group elements, their orders, and modular arithmetic.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that \( b^{\phi(n)} = 1 \mod(n) \) because \( \phi(n) \) represents the order of the group, suggesting that raising \( b \) to this power yields the identity element.
  • Another participant seeks clarification on the specific statement being proved, questioning whether \( b \) is an arbitrary element and the definitions of \( n \) and \( p \).
  • A third participant asks for clarification on the group being referenced, implying that the discussion may relate to Euler's and Fermat's theorems but noting that more details are needed.
  • A later reply specifies that the group in question is the multiplicative group modulo \( n \), explaining that raising \( b \) to the order of the group results in the identity element.
  • The same participant claims to prove the existence of the order of an element by demonstrating that for some powers of \( b \), \( b^m = b^n \) leads to \( b^{m-n} = e \), indicating that \( m-n \) is the order of \( b \).

Areas of Agreement / Disagreement

Participants express varying levels of understanding and clarity regarding the statements and proofs being discussed. There is no consensus on the specific proof or definitions being used, and multiple interpretations of the concepts are present.

Contextual Notes

Some assumptions about the definitions of \( n \), \( p \), and the nature of the group are not fully clarified, which may affect the understanding of the proofs. The discussion also reflects a dependence on the definitions of group order and modular arithmetic.

cragar
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I just want to make sure I understand these number theory proofs.
[itex]b^{\phi (n)}=1mod(n)[/itex]
[itex]\phi (n)[/itex] is the order of the group, so b to some power will equal the identity. so that's why it is equal to one. [itex]b^p=bmod(p)[/itex]
[itex]b^p=b^{p-1}b[/itex]
[itex]b^{p-1}[/itex] produces the identity since p-1 is the order of the group. so that's why it equals b.
 
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Can you clarify what statement you are trying to prove and which statements are the proof? Is b any arbitrary element of the group? Are you trying to show that p = Φ(n)? What is n?
 
And what group are you talking about?

I think you're trying to prove Euler and Fermat's little theorem somehow, but then you'll need to put in some more details...
 
Its the multiplicative group mod n , n is a natural number. now b is an element of the group. so b to power of the order of the group will contain the order of b.
that is [itex]b^m=e[/itex] since m is contained in the order of the group,we get the identity. This proves [itex]b^{\phi (n)}=1mod (n)[/itex].
the second one [itex]b^p=b mod (p) = b^{p-1}b=b^{\phi (p)}b= b mod (p)[/itex]
I will prove that the order of an element exists. sine b is an element of the group then we can take powers of b like [itex]b^2 , b^3, ... b^n ...[/itex] now for some m and n and m not equal to n,since our group is finite.
we get [itex]b^m=b^n[/itex] now we multiply both sides by [itex]b^{-n}[/itex]
so now [itex]b^{m-n}=e[/itex].
m-n is the order of b.
 
Last edited:

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