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Each equivalence class is a power of [g]

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
Hello!!! :)

I have to find an equivalence class $[g] \in \mathbb{Z_{15}}^{*}$ so that each equivalence class $\in \mathbb{Z}^{*}_{15}$ is a power of $[g]$.

$\mathbb{Z}^{*}_{15}=\{[1],[2],[4],[7],[8],[11],[13],[14]\}$

I tried several powers of the above classes,and I think that there is no equivalence class $[g] \in \mathbb{Z_{15}}^{*}$ so that each equivalence class $\in \mathbb{Z}^{*}_{15}$ is a power of $[g]$.Is it actually like that or am I wrong?? (Thinking)
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Re: each equivalence class is a power of [g]

In fact:

$\langle [1]\rangle = \{[1]\}$

$\langle [2]\rangle = \{[1],[2],[4],[8]\} = \langle [8]\rangle$

$\langle [4]\rangle = \{[1],[4]\}$

$\langle [7]\rangle = \{[1],[7],[4],[13]\} = \langle [13]\rangle$

$\langle [11]\rangle = \{[1],[11]\}$

$\langle [14]\rangle = \{[1],[14]\}$

which shows that every element has order 1,2 or 4, and that no element has order 8.

(for $g > 7$ it is easier to compute $\langle[g]\rangle$ as $\langle[-(15-g)]\rangle$).