Showing $(\zeta^{i-1})^{-1}\circ(1\, j+1)\circ\zeta^{i-1}=(i \, i+j)$.

Question

This is part of Exercise 1.9.6 of Howie's "Fundamentals of Semigroup Theory". Here composition is from left to right.

Let $n\in\mathbb{N}$ and let $\zeta=(12\dots n)$ (a cycle in the symmetric group $S_n$). Show that, for $i\in\{1, \dots , n-1\}$ and $j\in\{1, \dots , n-i\}$, $$(\zeta^{i-1})^{-1}\circ(1\, j+1)\circ\zeta^{i-1}=(i \, i+j).$$

My Attempt:

I've tried induction on $j$ with $i$ fixed but I haven't got anywhere. The base case is a previous part, which states that $$(\zeta^{i-1})^{-1}\circ (12)\circ\zeta^{i-1}=(i\, i+1).$$ This is proven by induction on $i$.

Answer 1

$\sigma^{-1}(ab)\sigma$ is always $(a' b')$ where $a',b'$ are the images of $a,b$ under $\sigma$. In this case clearly the image of $1$ is $i$ and the image of $j+1$ is $j+i$.