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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

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1 Answer

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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.

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