I am trying to go through some lecture notes found here, and am struggling with an assertion on page 21. It is meant to be a proof that any cycle $(a_1,..,a_k)$ can be written as a product of transpositions.
We have the identity $(1,k)(1,k-1)...(1,3)(1,2) = (1,2,...,k)$.
Now consider a permutation $\phi$ taking $1 \rightarrow a_1, ..., k\rightarrow a_k$. Using the above identity, we can write
$(a_1,..,a_k) = \phi \circ(1,k)...(1,2)\circ \phi^{-1}$.
The claim is that this is our desired decomposition. We have not shown that $\phi$ itself can be written as transpositions, so how do I see that the right-hand-side of the above equation is indeed just some transpositions?
EDIT: As pointed out in the comments, we just have that $$ \begin{split} (a_1,..,a_k) &= \phi \circ(1,k)...(1,2)\circ \phi^{-1} \\ &= \phi \circ(1,k)\circ\phi^{-1} \circ \phi\circ(1,k-1)...(1,2)\circ \phi^{-1} \\ &= (a_1,a_k)(a_1,a_{k-1})...(a_1,a_2), \end{split} $$ giving the result.