How to Determine the Order of \( g^8 \) in a Group?

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Discussion Overview

The discussion centers on determining the order of the element \( g^8 \) in a group where the element \( g \) has an order of \( 28 \). The focus is on mathematical reasoning and exploring methods to find the order of powers of elements in group theory.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant suggests finding the order of \( g^8 \) by determining the lowest \( k \) such that \( (g^{8})^k = 1 \), noting that \( 28 \) is the order of \( g \).
  • Another participant reiterates the approach of finding \( k \) such that \( 8k \) is a multiple of \( 28 \) and proposes using the least common multiple (lcm) to express \( k \) as \( k = \frac{\text{lcm}(28,8)}{8} \).
  • A later reply questions whether there is a systematic method for finding the order of \( g^i \) for all \( 2 \le i \le 27 \), suggesting an alternative formula \( k = \frac{28}{\gcd(28, i)} \) based on common prime factors.
  • Participants express uncertainty about the systematic nature of the proposed methods, indicating that the approach may not be comprehensive.

Areas of Agreement / Disagreement

Participants generally agree on the methods to find the order of \( g^8 \) and propose alternative approaches, but there is no consensus on a comprehensive systematic method for all powers \( g^i \).

Contextual Notes

The discussion does not resolve the completeness of the proposed methods or their applicability to all cases, leaving some assumptions about the nature of the group and the elements involved unaddressed.

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Say an element $g$ in a group has order $28$. How do I find the order of say $g^8$?
 
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Guest said:
Say an element $g$ in a group has order $28$. How do I find the order of say $g^8$?

Hi Guest,

We are looking for the lowest $k$ such that $(g^{8})^k = 1$.
And we know that $28$ is the lowest such that $g^{28} = 1$.
That means we're looking for the lowest $k$ such that $8k$ is a multiple of $28$.
That is:
$$k = \frac{\text{lcm}(28,8)}{8}$$
 
I like Serena said:
Hi Guest,

We are looking for the lowest $k$ such that $(g^{8})^k = 1$.
And we know that $28$ is the lowest such that $g^{28} = 1$.
That means we're looking for the lowest $k$ such that $8k$ is a multiple of $28$.
That is:
$$k = \frac{\text{lcm}(28,8)}{8}$$
Thanks. I wonder whether there's a systematic way of working this out if one has to find $g^i$ for all $2 \le i \le 27$?
 
Last edited:
Guest said:
Thanks. I wonder whether there's a systematic way of working this out if one has to find $g^i$ for all $2 \le i \le 27$?

Alternatively, we can write it as:
$$k=\frac{28}{\gcd(28, i)}$$
That is, find the common prime factors and divide 28 by them.

I'm afraid that's as systematic as it gets.
 

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