Kiwi1
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Prove:
Let \alpha = (a_1,...,a_s) be a cycle and let \pi be a permutation in Sn. Then \pi \alpha \pi ^{-1} is the cycle (\pi(a_1), ... \pi(a_s))
My attempt.
(\pi \alpha \pi ^{-1})^s = (\pi \alpha^s \pi ^{-1})=e so if this thing is a cycle and its length divides s.
Assume \pi (a_1) is a member of the cycle. Then:
\pi \alpha \pi ^{-1}(\pi (a_1))=\pi (a_2)
and
\pi \alpha \pi ^{-1}(\pi (a_i))=\pi (a_{i+1}) for 1 \leq i < s
and finally
\pi (a_{s+1})=\pi \alpha \pi ^{-1}(\pi (a_s))=\pi(a_1) (validating the assumption)
I think I have all of the components of a proof here. But how can I make it more rigorous?
Let \alpha = (a_1,...,a_s) be a cycle and let \pi be a permutation in Sn. Then \pi \alpha \pi ^{-1} is the cycle (\pi(a_1), ... \pi(a_s))
My attempt.
(\pi \alpha \pi ^{-1})^s = (\pi \alpha^s \pi ^{-1})=e so if this thing is a cycle and its length divides s.
Assume \pi (a_1) is a member of the cycle. Then:
\pi \alpha \pi ^{-1}(\pi (a_1))=\pi (a_2)
and
\pi \alpha \pi ^{-1}(\pi (a_i))=\pi (a_{i+1}) for 1 \leq i < s
and finally
\pi (a_{s+1})=\pi \alpha \pi ^{-1}(\pi (a_s))=\pi(a_1) (validating the assumption)
I think I have all of the components of a proof here. But how can I make it more rigorous?