We have the group $(S_3, \circ)$, where $S_3=\{(1)(2)(3),(1)(2,3),(1,2)(3),(1,2,3),(1,3,2),(1,3)(2)\}$. How can I find all cyclic subgroups from this?
I know that I could go through each element and exponentiate it to see if I get all other terms. However, I would guess there is a more efficient way of solving this problem.
Best Answer
The order of an element must divide the order of the group. The order of $S_3$ is $6$, and $S_3$ is not cyclic; that leaves $1, 2, $ and $3$ as possible orders for elements of $S_3$. The cyclic group of order $1$ has just the identity element, which you designated $(1)(2)(3)$. The order of an element in a symmetric group is the least common multiple of the lengths of the cycles in its cycle decomposition. Therefore, the orders of $(1)(2,3), (1,2)(3), $ and $(1,3)(2)$ are $2$, and the orders of $(1,2,3)$ and $(1,3,2)$ are $3$. [The elements of order $2$ are self-inverses. $(1,2,3)$ and $(1,3,2)$ are inverses of each other.] $S_3$ is a good example group to study, because it is the smallest non-Abelian group.