Simplifying permutations written in cycle notation

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I was wondering how I could simplify $((1 3 5)(2 3 4 1)(3 2 1))^{−1}$.

I was going to simplify the inner bracket section first then apply the inverse but I am having trouble simplifying it.

I see that $1\to3\to5\to1 $ but also $1\to3\to4\to1$ and$1\to3\to2\to1$. So I am not sure how to simplify it into one expression?

Thank you, if anyone could help that would be great :)

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$$\begin{align*}(135) \text{ takes }1 \to 3 \\ (2341)\text{ takes }3 \to 4 \\ (321) \text{ takes }4 \to 4\end{align*}$$

So, (135)(2341)(321) takes $1 \to 4$, so the inverse would take $4 \to 1$.

$$\begin{align*}(135)\text{ takes }2 \to 2 \\ (2341)\text{ takes }2 \to 3 \\ (321)\text{ takes }3 \to 2\end{align*}$$

So, (135)(2341)(321) takes $2 \to 2$, so the inverse would take $2\to 2$.

Etc.

Final result: $$(135)(2341)(321) = (1435)$$

$$((135)(2341)(321))^{-1} = (1534)$$

Edit: if the author does multiplication right to left:

$$\begin{align*}(321)\text{ takes }1 \to 3 \\ (2341)\text{ takes }3\to 4 \\ (135) \text{ takes }4 \to 4\end{align*}$$

etc.

Final Result:

$$(135)(2341)(321) = (1435)$$

$$((135)(2341)(321))^{-1} = (1534)$$

This problem happens to work out the same if it is left-to-right or right-to-left multiplication (not all problems will work out that way).