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Recently I asked how to calculate the order of an element in $U_{27}=\mathbb{Z}_{27}^*$ (Multiplicative group of integers modulo $27$) (link). Problem is I still don't understand the material but I would like to explain what I know and what I don't. Tried to find some similar thread on the same topic and I found the following tread (link). Although it does not answer my question directly, it points into that direction. I know the hard way to find the order of an element in $U_{27}$. For example in order to find the order of $5$ in $U_{27}$ I would do: $$\begin{align*} 5&=5\bmod 27=5\\ 5^2&=25\bmod 27=25\\ 5^3&=125\bmod 27=17 \end{align*}$$

And so on, until I find $n\in\mathbb{N}$ so $5^n=1$. It could take awhile, in fact I know that the order of $5$ in $U_{27}$ is $18$ ($5^{18}=3814697265625(mod27)=1$), so I will have to calculate $18$ times and facing some big numbers. I think that there is a fast way using the euler function. How can I use it in my advantage? is there a formula?

Chinnapparaj R
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vesii
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    Hint: If you know $5^3 \equiv 125 \mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 \mod 27$? – David G. Stork Nov 20 '18 at 20:43
  • $|\mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1. – user614287 Nov 20 '18 at 20:46

2 Answers2

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First of all note that,

  • if $G$ is group and $a\in G$ then $o(a)=o(a^{-1})$.

  • if $G$ is group and $a\in G$ be an element of order $m$ then $o(a^k)=\frac{m}{gcd(m,k)}$ where $k\in \mathbb{N}$

    By using these two results you can avoid half of calculation needed to compute order of elements.

Now let's start,

$o(1)= 1$ since $1$ is identity in $U_{27}$.

$o(2)= 18$ since $2^{18}≡1\mod 27$ and $2^m ≢ 1\mod 27$ for $m<18 \in \mathbb{N}$. Now using our above results:

$o(4)=o(2^2)=\frac{o(2)}{gcd(o(2),2)}=\frac{18}{gcd(18,2)}=\frac{18}{2}=9$

$o(8)=o(2^3)= \frac{o(2)}{gcd(o(2),3)}=\frac{18}{gcd(18,3)}=\frac{18}{3}=6$

Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)

To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$

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Akash Patalwanshi
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Here $U_{27}$ is a cyclic group of order $18=2 \times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k \geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]

Here $\langle5 \rangle= U(9) $ and so $\langle 5 \rangle= U(3^k) ,k \geq 2$ and in particular $\langle 5 \rangle= U(3^3)=U(27) $, so $\vert 5 \vert =18$

Chinnapparaj R
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