Prove for $\sigma \in S_n$, $\sigma(a_1, … , a_n)\sigma^{-1}= (\sigma(a_1),…, \sigma(a_n))$

Question

How do I even begin this problem? Does it involve induction or something like induction?

Would it be more helpful to instead prove it as $$\sigma(a_1,..., a_n) = (\sigma(a_1),...,\sigma(a_m))\sigma\ ?$$ I'm lost.

Answer 1

It suffices to show that for any $i$, $$ [\sigma(a_1,a_2,\dots,a_n) \sigma^{-1}](\sigma(a_i)) = \sigma(a_{i+1}) $$ where the indices $i$ are taken modulo $n$. No need for induction, just plug in and simplify.