MHB Orders of Elements in Groups $$\Bbb{Z}_{12}, U(10), U(12), D4$$

[karush]

nmh{909}
For each group in the following list,
$$ \Bbb{Z}_{12}, \qquad U(10)\qquad U(12) \qquad D4 $$
(a) find the order of the group
$$|\Bbb{Z}_{12}|=12$$
(b) the order of each element in the group.ok the eq I think we are supposed to use is
$$\textit{ if } o(g)=n \textit{ then } o(g^n)= n/(n,k)$$
the alleged a answer for (a) is $\Bbb{Z}_{12}$
for (b)$o(0)=1, \quad $o(1)=12$ \quad $o(2)=6$\quad $o(3)=4$\quad $o(4)=3$,\quad $o(5)=12$
\quad $o(6)=2$\quad $o(7)=12$\quad $o(8)=3$\quad $o(9)=4$\quad $o(10)=6$\quad $o(11)=12$I am sure this is simple but don't see it

 

[I like Serena]

Hi karush,

The order an element in $\mathbb Z_{12}$is the number of times we need to add the element before we get 0 modulo 12.
For instance $(4+4+4) \bmod{12}=12 \bmod{12}=0$.
So the order of $4$ is $o(4)=3$.

 

[karush]

Hi karush,
The order an element in $\mathbb Z_{12}$is the number of times we need to add the element before we get 0 modulo 12.
For instance $(4+4+4) \bmod{12}=12 \bmod{12}=0$.
So the order of $4$ is $o(4)=3$.

so then
$o(1)=12$ is $(1+1+1+1 +1+1+1+1 +1+1+1+1) \bmod{12}=0$
and
$o(2)=6$ is $(2+2+2+2+2+2+2) \bmod{12} =0$

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