Order of an element in ##\mathbb{Z}_n##

[chwala]

TL;DR Summary

See attached...

Doing some self study here; my understanding of order of an element in a group is as follows:
Order of ##3## in ##\mathbb{z_4}## can be arrived by having, ##3+3+3+3=12≡0##
likewise, the order of ##12## in ##\mathbb{z_{20}}## can be arrived by
##12+12=24 ≡4≠0##
##12+12+12=36≡16≠0##
##12+12+12+12=48≡18≠0##
##12+12+12+12+12=60≡0⇒12^5=0##
therefore the order of ##12## in ##\mathbb{z_{20}}=5##. Any insight is welcome.

 

[pasmith]

Quicler is, for positive integers [itex]n[/itex] and [itex]m[/itex], [tex]
12n - 20m = 4(3n - 5m) = 0\quad\Leftrightarrow\quad (n,m) = (5k,3k), k \in \mathbb{N}[/tex] and the order of 12 in [itex]\mathbb{Z}_{20}[/itex] is found by taking [itex]k = 1[/itex].

 

[chwala]

...This is a continuation

This part is quite clear... that is:
##3## in ##\mathbb{z_4}=4##
##6## in ##\mathbb{z_{12}}=2##
##12## in ##\mathbb{z_{20}}=5##
##16## in ##\mathbb{z_{24}}=3##
and
lcm ##(4,2,5,3)=60##

 

[chwala]

Quicler is, for positive integers [itex]n[/itex] and [itex]m[/itex], [tex]
12n - 20m = 4(3n - 5m) = 0\quad\Leftrightarrow\quad (n,m) = (5k,3k), k \in \mathbb{N}[/tex] and the order of 12 in [itex]\mathbb{Z}_{20}[/itex] is found by taking [itex]k = 1[/itex].

Looks easy to apply, then for
##16## in ##\mathbb{z_{24}}## we shall have,
##16n-24m=8(2n-3m)=0## ...
##(3k,2k)##, on taking ##k=1## we shall then have the required ##n=3##.

 

[chwala]

TL;DR Summary: See attached...

View attachment 330097

Doing some self study here; my understanding of order of an element in a group is as follows:
Order of ##3## in ##\mathbb{z_4}## can be arrived by having, ##3+3+3+3=12≡0##
likewise, the order of ##12## in ##\mathbb{z_{20}}## can be arrived by
##12+12=24 ≡4≠0##
##12+12+12=36≡16≠0##
##12+12+12+12=48≡18≠0##
##12+12+12+12+12=60≡0⇒12^5=0##
therefore the order of ##12## in ##\mathbb{z_{20}}=5##. Any insight is welcome.

Let me have this here for future reference...my way of attempting,
##\mathbb{z_3} ×\mathbb{z_4}=(0,1,2) ×(0,1,2,3)##

Giving us,

##(0,0), (0,1), (0,2), 0,3)##
## (1,0), (1,1), (1,2), (1,3)##
##(2,0), (2,1), (2,2), (2,3)##
the order is ##12##.

 

[chwala]

For ##\mathbb{z_2} ×\mathbb{z_2}×\mathbb{z_2}## ...i am going through this link

https://quizlet.com/explanations/qu...ups-of-2-e8129f84-d882-40fe-baab-e8c7b00823ab

My way of doing it,

##\mathbb{z_2} ×\mathbb{z_2}=(0,1)×(0,1)=(0,0), (0,1), (1,0), (1,1)##

...

##\mathbb{z_2} ×\mathbb{z_2}×\mathbb{z_2}=[(0,0), (0,1), (1,0), (1,1)]×[(0,1)]##

##=(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)## and ##(1,1,1)##

order is ##8##.
Insight is welcome.

 

[fresh_42]

For ##\mathbb{z_2} ×\mathbb{z_2}×\mathbb{z_2}## ...i am going through this link

https://quizlet.com/explanations/qu...ups-of-2-e8129f84-d882-40fe-baab-e8c7b00823ab

My way of doing it,

##\mathbb{z_2} ×\mathbb{z_2}=(0,1)×(0,1)=(0,0), (0,1), (1,0), (1,1)##

You should distinguish elements and sets.
$$\mathbb{Z}_2 \times\mathbb{Z}_2=\{0,1\}\times \{0,1\}=\{(0,0), (0,1), (1,0), (1,1)\}$$
and write these cyclic groups with a capital Z, and the index outside of the brackets:

\mathbb{Z}_p

...

##\mathbb{z_2} ×\mathbb{z_2}×\mathbb{z_2}=[(0,0), (0,1), (1,0), (1,1)]×[(0,1)]##

Yes. We have three direct factors here, which are triplets, so
\begin{align*}
\mathbb{Z}_2^3&=\mathbb{Z}_2 \times\mathbb{Z}_2\times\mathbb{Z}_2\\&=\{0,1\}\times \{0,1\}\times \{0,1\}\\&=\{(0,0,0), (0,1,0), (1,0,0), (1,1,0),(0,0,1), (0,1,1), (1,0,1), (1,1,1)\}
\end{align*}
These direct products are associative so it does not matter where or even whether you insert an order.

##=(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)## and ##(1,1,1)##

order is ##8##.
Insight is welcome.

Edit: "You should distinguish elements and sets."

I should do that, too. I corrected the notation (from ##\{(0,1)\}## to ##\{0,1\}## for the single group ##\mathbb{Z}_2##).

 

[Mark44]

You should distinguish elements and sets.

and write these cyclic groups with a capital Z, and the index outside of the brackets:

Yes to both...