Let's say G is the symmetric group of degree 5. Suppose I wanted to find the cyclic subgroup generated by the permutation $x = (1,2,3)(4,5)$.
I know that the order of x is 6 – so I know the cyclic subgroup will have 6 elements.
To find the cyclic subgroup, would it be sufficient to simply find $x^0, x^1, x^2, x^3, x^4, x^5, x^6$ – is this set guaranteed to be the cyclic subgroup generated by x? And is this always the case?
Thanks for any help,
Jack
Best Answer
The cyclic subgroup generated by an element $x$ is by definition the set of powers $$\{x^k|k\in\mathbb Z\}$$ If $x$ is of finite order $n$, this is just $$\{x^0,\ldots,x^{n-1}\}=\{x^1,\ldots,x^n\}$$ You can actually prove this. Give it a shot.