Browse Questions

# Find the order of each element in the group $(Z_5 − \{[0]\}, _.5)$

Toolbox:
• Multiplication modulo n $(._{\large n}):a._{\large n} b=r ; 0\leq r< n$,where r is the least non-negative remainder when $ab$ is divided by n.
• Congruence modulo n :Let $a, b \in Z$ and $n$ be a fixed positive integer.
• We say that “a is congruent to b modulo n” $\Leftrightarrow (a − b)$ is divisible by n Symbolically,
• $a \equiv b (mod\; n) \Leftrightarrow (a − b)$ is divisible by n.
• Let G be a group and $a \in G$. The order of ‘a’ is defined as the least positive integer n such that $a^n = e$, $e$ is the identity element. If no such positive integer exists, then a is said to be of infinite order. The order of a is denoted by 0(a).
• Here $a^n = a * a * a ... *a$ (n times). If $*$ is usual multiplication ‘.’ then $a^n$ is $a . a .a...$(n times) i.e., $a^n .$
Step 1:
Consider the group $(z_5-\{[0]\},._5)$
$z_5-\{[0]\}=\{[1],[2],[3],[4]\}$
Step 2:
To find the order of each of the elements under $._5[1]$ is the identity element under modular multiplication.
$\therefore 0([1])=1$
$[2]^2=[4],[2]^3=[4]._5[2]=[3],[2]^4=[3]._5[2]=[1]$
$\therefore 0([2])=4$
$[3]^2=[4],[3]^3=[4]._5[3]=[2],[3]^4=[2]._5[3]=[1]$
$\therefore 0([3])=4$
$[4]^2=[1]$
$\therefore 0([4])=1$
We have $0([1])=1,0([2])=4,0([3])=4,0([4])=2$