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Find the orders of [5], and [12] in the additive group modulo 18?
I want know if what I am doing is correct. Here is my attempt
[5] + [5] =[10]
[5] + [5] + [5]= [15]
[5]+ [5] + [5] + [5]= [20]=[2]
based on these few steps, do I use this process? I came up with 18 when using [5]. Am I right?
If you need definitons, just ask.

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    Since $5$ and $18$ are coprime, $5$ _generates_ the group (recall Bezout's identity), so its order must be $18$. – 2017-01-22
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    yeah but the process takes a while based on my notes. Somebody just gave me simpler way – 2017-01-22

2 Answers 2

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it looks fine, although there is a shortcut.

The order of $n$ in the additive group $\mathbb Z_n$ is $\frac{18}{gcd(n,18)}$. So the order of $5$ is $\frac{18}{1}$ and the order of $12$ is $\frac{18}{6}=3$.

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    so use the gcd to make it simple? Your way seems much quicker.Thanks – 2017-01-22
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    Sure no problem. – 2017-01-22
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    can you give me the simply way for the multiplicative form also? I need to find its order of [4] in the group modulo 7. Its tedious doing it by hand in the multiplicative form – 2017-01-22
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    @behold sadly no such easy formula exists for the multiplicative order – 2017-01-22
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    so I did the multiplicative of [4] modulo 7 and I got $[4]^3=[64]=[1]$. I think that is the correct way to do it – 2017-01-22
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    yup, the answer is always a divisor of $\varphi(n)$ by the way (where $n$ is the modulus). – 2017-01-22
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    Okay. I am on the right track but $[4]^4$ is giving me trouble. I got 4 as an answer for that – 2017-01-22
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The order of $a$ modulo $18$ is the smallest $r$ such that $ra$ is a multiple of $18$, i.e. is a multiple of both $18$ and $a$. In other words, $ra$ is the l.c.m. of $a$ and $18$.

Now we know that $\;\operatorname{lcm}(a,18)\times \gcd(a,18)=a\times 18$, whence $$\operatorname{lcm}(a,18)=\frac{a}{\gcd(18,a)}\times18,\enspace\text{so}\quad r=\frac{a}{\gcd(18,a)}.$$