0
$\begingroup$

This is a only theoritical.

Why is the order $o( \bar x )$ of $\bar x∈\mathbb Z_n$ the smallest non-negative integer $k$ such that $kx \equiv 0$ (mod $n$)?

I don't understand how it follows from the general definition of orders in group theory. Maybe I don't even get what $\bar x$ refers to but I think I do.

Can you help me? If you can tell me what you think I don't understand or just explain to me, it would of great help! Thank you

$\endgroup$
2
  • $\begingroup$ Typically in this notation $x$ refers to an element of $\mathbb Z$, and $\overline x$ refers to the coset represented by $x$ in $\mathbb Z/n\mathbb Z$. Note that $\overline{kx} = \overline x + \cdots + \overline x$, with $k$ copies of $\overline x$ on the RHS. Now can you see the connection to the definition of the order of $\overline x$? $\endgroup$ – Dustan Levenstein Apr 19 '14 at 8:42
  • $\begingroup$ Hm... x¯ is a set then? Can you give me an example introducing modular arithmetic? I think I'd see more clearly but I can't find any on the web. $\endgroup$ – Timmy Apr 19 '14 at 9:38
2
$\begingroup$

A quick answer..

For a fixed positive integer $n$, An equivalence relation ~ on $\mathbb Z$, The ring of Integers, defined by $a$~$b$ $\Leftrightarrow n$ devides $a-b$ , partitions $\mathbb Z$ into $n$ equivalence classes $\bar 0, \bar 1, \cdots ,\overline{n-1}$ $\quad$ where $\bar r=\{ nk+r: k\in \mathbb Z\}$, $0\leq r\leq n-1$.

The quotient set $\mathbb Z_n=\{ \bar 0, \bar 1, \cdots ,\overline{n-1}\}$ forms an abelian (moreover cyclic) group with respect to the binary operation $\oplus :\mathbb Z_n \times\mathbb Z_n\longrightarrow \mathbb Z_n$ defined by $\bar a\oplus \bar b =\overline{a+b}$.

Order of an element $g$ of a group $G$ is $\min\{m\in \mathbb N : g^m=1_G\}$.

So, order $o(\bar x)$ of $\bar x\in \mathbb Z_n$ is $\min\{m\in \mathbb N : \underbrace{\bar x\oplus \bar x\oplus \cdots \oplus \bar x}_{m ~ ~ times}=\overline{mx}=\bar0 \}=\min\{ m\in \mathbb N: mx\equiv 0 ~ (mod ~ n)\}$.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! That helps more than you imagine! $\endgroup$ – Timmy Apr 19 '14 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.