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

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SUMMARY

The order of the element \( g^8 \) in a group where \( g \) has an order of \( 28 \) can be determined by finding the lowest integer \( k \) such that \( (g^8)^k = 1 \). This requires calculating \( k \) as \( k = \frac{\text{lcm}(28, 8)}{8} \). Additionally, for any \( g^i \) where \( 2 \leq i \leq 27 \), the order can be systematically found using the formula \( k = \frac{28}{\gcd(28, i)} \), which involves identifying the common prime factors of \( 28 \) and \( i \).

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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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