For instance, if I take the permutation $\alpha=(123)(67)(458) \in S_{10}$, what is the subgroup generated by it? Knowing that the order of $\alpha$ is 6, I already calculated $\alpha ^0, \alpha ^1$ and so on, until $\alpha ^5$. How do I proceed from this?
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It seems you have calculated every element of the subgroup then. Namely $\alpha^0,\alpha^1,\alpha^2,\alpha^3,\alpha^4, \alpha^5$. – Harto Saarinen Nov 23 '15 at 16:20
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Is that all I need to do? – Encontro Regional Sul Nov 23 '15 at 16:21
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1Yes those all the elements that form the subgroup generated by $\alpha$. You can check that they really form a subgroup if you want. – Harto Saarinen Nov 23 '15 at 16:23
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Because your permutation has order $6$ you have found all the elements of subgroup generated by $\alpha$. The subgroup is $\langle \alpha \rangle = \{\alpha^0 = id, \alpha^1, \alpha^2, \alpha^3, \alpha^4, \alpha^5\}$. It's an easy exercise (if you don't see it right away) to verify this really is a subgroup.
It's maybe worth to check that your permutation really has order 6. You can see that noting that $lcm(3,2,3) = 6$ (Those numbers $3,2,3$ are length of each cycle in $\alpha$).
Harto Saarinen
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Thank you. And yeah, I used the least common divisor result to say that $\alpha$ has order 6. – Encontro Regional Sul Nov 23 '15 at 16:54