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

  • $\begingroup$ 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

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

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

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