4 votes 4 votes A={1,2,3,4,5,6} Given that set A is a group with respect to multiplication mod 7.Is A a cyclic group?If yes,identify the generators. Is there any theorem that i cana pply here to solve without building operation table? Set Theory & Algebra discrete-mathematics group-theory + – rahul sharma 5 asked Feb 3, 2017 rahul sharma 5 3.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 9 votes 9 votes $\begin{align*} &\text{A finite group G is CYCLIC} \Leftrightarrow \exists a \in \text{G such that O(a) = O(G)} \\ &\text{Here O(G) = 6} \\ &\text{So we can check for each } {\bf a_i} \in \text{G whether } { a_i^{6} = 1} \text{ or not ?} \\ \\ \hline \\ &\text{Start with id which is 1} \rightarrow \text{Not a gen} \;\;\; \left \{\bf \text{if }O(G) > 1 \rightarrow \text{ id }\neq\text{ gen} \right \} \\ &\text{Next 2: } \\ &\begin{matrix} &2^0&={\color{green}{1}}\\ &2^1&={\color{green}{2}}\\ &2^2&={\color{green}{4}}\\ &2^3&={\color{red}{1}}&\text{repeated [stop]}\\ \end{matrix}\\ &\text{Next 3: } \\ &\begin{matrix} &3^0&={\color{green}{1}}\\ &3^1&={\color{green}{3}}\\ &3^2&={\color{green}{2}}\\ &3^3&={\color{green}{6}}\\ &3^4&={\color{green}{4}}\\ &3^5&={\color{green}{5}}\\ &3^6&={\color{green}{1}}&\text{all generated [stop]} \\ \end{matrix}\\ &\Rightarrow 3 \text{ is a generator}\\ &\Rightarrow \text{ G is cyclic} \\ \end{align*}$ $\begin{align*} &\Rightarrow 3^{-1} \text{ is also a generator which is 5} \\ \\ \hline \\\\ &\text{Once we figured out that } \text{ G } \text{ is cyclic with O(G) = 6}\\ &\text{Then } {\bf \phi(6)} \text{ will give no of generators in G} \\ &\Rightarrow {\bf \phi(6)} = {\bf \phi(2^1*3^1)} \\ &\Rightarrow {\bf \phi(6)} = {\bf \phi(2^1)}*{\bf \phi(3^1)} \\ &\Rightarrow {\bf \phi(6)} = \left [ 2^1-2^0 \right ]*\left [ 3^1-3^0 \right ] \\ &\Rightarrow {\bf \phi(6)} = 2 \\ &\Rightarrow \text{ And these two gen are 3 and 5} \\ \end{align*}$ dd answered Feb 3, 2017 selected Feb 3, 2017 by Rahul Jain25 dd comment Share Follow See all 0 reply Please log in or register to add a comment.
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. Vijay Thakur answered Feb 3, 2017 Vijay Thakur comment Share Follow See all 2 Comments See all 2 2 Comments reply rahul sharma 5 commented Feb 3, 2017 reply Follow Share Your answer is correct.But order of the group represents number of elements?Here number of elements are 6,so how it will be prime?Am i missing something? And how did u find generators? 1 votes 1 votes sudsho commented Feb 3, 2017 reply Follow Share ^ he has answered without the loss of generality here..means in general if order of group is prime than that will be cyclic....for this question u cant apply this rule...so try to find a generator... take each element and try to derive all others...u'll see that 3 can derive all others in the group(like 36 is 1, 32 is 2 and so on...) inverse of a generator is also a generator..means 5.. also one property is that order of generator is the order of the group....like 36=1 which is the identity element..so 6 is the order of the group... 2 votes 2 votes Please log in or register to add a comment.