Multiplication of cyclic permutation

Question

Find $f^{-1}gf$, where $f=(123)$, $g=(2345)$.

Since $f$ and $g$ are cyclic permutations, we have $gf=fg$. Hence $f^{-1}gf=f^{-1}fg= (f^{-1}f)g=ig=g=(2345)$, where $i$ is identity permutation. Since i am new to this particular topic, i am not sure whether i am doing correctly?

Answer 1

It's not true that cycles commute always, disjoint cycles do.

So you just have to do the computation the standard way, tracing images:

$$(123)^{-1} (2345) (123)= (321)(2345)(123) = (1245)$$

Answer 2

One can prove that $$\forall \sigma \in \mathfrak S_n, \ \ \sigma (i_1 ...i_m)\sigma ^{-1}=(\sigma (i_1)...\sigma (i_m)),$$ where $i_1,...,i_m\in \{1,...,n\}$ are distincts.