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

[chwala]

TL;DR

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 n and m, <br /> 12n - 20m = 4(3n - 5m) = 0\quad\Leftrightarrow\quad (n,m) = (5k,3k), k \in \mathbb{N} and the order of 12 in \mathbb{Z}_{20} is found by taking k = 1.

 

[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 n and m, <br /> 12n - 20m = 4(3n - 5m) = 0\quad\Leftrightarrow\quad (n,m) = (5k,3k), k \in \mathbb{N} and the order of 12 in \mathbb{Z}_{20} is found by taking k = 1.

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...