Show that $σ(i_1, i_2, . . . , i_k)σ^{−1} = (σ(i_1), σ(i_2), . . . , σ(i_k))$

Question

Here's the full question: If $σ ∈ S_n$ is any permutation and $i_1, . . . , i_k $ are $k$ distinct elements of $\{1, . . . , n\}$, show that $σ(i_1, i_2, . . . , i_k)σ^{−1} = (σ(i_1), σ(i_2), . . . , σ(i_k))$

I was given this hint:

For $j ∈ \{1, . . . , n\}$ consider these cases: (1) $j$ is one of the values $σ(i_1), . . . ,σ(i_{k−1})$, (2) $j = σ(i_k)$, (3) $j$ is not one of the values $σ(i_1), . . . , σ(i_k)$.

I'm supposed to show that both sides send $j$ to the same thing, but the wording of this question confused me and the hint only worsened this for me.

Answer 1

See the solution to Exercise 5.17 (a) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Note that your notation $\left(i_1, i_2, \ldots, i_k\right)$ corresponds to my notation $\operatorname{cyc}_{i_1,i_2,\ldots,i_k}$.

At some point you'll get used to this kind of proof, and begin saying that they are obvious; right now, probably having all the details available is a good thing.