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Best answer
9 votes
9 votes
$\begin{align*} &\text{A finite group G is CYCLIC} \Leftrightarrow \exists a \in \text{G such that O(a) = O(G)} \\ &\text{Here O(G) = 6} \\ &\text{So we can check for each } {\bf a_i} \in \text{G whether } { a_i^{6} = 1} \text{ or not ?} \\ \\ \hline \\ &\text{Start with id which is 1} \rightarrow \text{Not a gen} \;\;\; \left \{\bf \text{if }O(G) > 1 \rightarrow \text{ id }\neq\text{ gen} \right \} \\ &\text{Next 2: } \\ &\begin{matrix} &2^0&={\color{green}{1}}\\ &2^1&={\color{green}{2}}\\ &2^2&={\color{green}{4}}\\ &2^3&={\color{red}{1}}&\text{repeated [stop]}\\ \end{matrix}\\ &\text{Next 3: } \\ &\begin{matrix} &3^0&={\color{green}{1}}\\ &3^1&={\color{green}{3}}\\ &3^2&={\color{green}{2}}\\ &3^3&={\color{green}{6}}\\ &3^4&={\color{green}{4}}\\ &3^5&={\color{green}{5}}\\ &3^6&={\color{green}{1}}&\text{all generated [stop]} \\ \end{matrix}\\ &\Rightarrow 3 \text{ is a generator}\\ &\Rightarrow \text{ G is cyclic} \\ \end{align*}$

$\begin{align*} &\Rightarrow 3^{-1} \text{ is also a generator which is 5} \\ \\ \hline \\\\ &\text{Once we figured out that } \text{ G } \text{ is cyclic with O(G) = 6}\\ &\text{Then } {\bf \phi(6)} \text{ will give no of generators in G} \\ &\Rightarrow {\bf \phi(6)} = {\bf \phi(2^1*3^1)} \\ &\Rightarrow {\bf \phi(6)} = {\bf \phi(2^1)}*{\bf \phi(3^1)} \\ &\Rightarrow {\bf \phi(6)} = \left [ 2^1-2^0 \right ]*\left [ 3^1-3^0 \right ] \\ &\Rightarrow {\bf \phi(6)} = 2 \\ &\Rightarrow \text{ And these two gen are 3 and 5} \\ \end{align*}$
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2 votes
2 votes
As far as I can remember a group with prime number order is cyclic.

I think 3 is generater and its inverse  5 is generator too.

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