How does $(abc) = (ac)(ab)$? I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots  \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$
However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.
Thank you in advance!

We can also conveniently calculate the product of transpositions in cycle notation by switching to the two-line notation of permutations.
Considering the first identity $(abc)=(ac)(ab)$ we have the representation
\begin{align*}
(abc)\equiv\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\qquad\qquad&
(ac)\equiv\begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\\
&(ab)\equiv\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
\end{align*}

We obtain
\begin{align*}
\color{blue}{(ac)(ab)}&\equiv \begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\circ\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
&=\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\tag{1}\\
&\,\,\color{blue}{\equiv(abc)}\tag{2}
\end{align*}

Comment:

*

*In (1) we use the convention to multiply out from right to left:
\begin{align*}
&a\to b\to b\\
&b\to a\to c\\&
c\to c\to a
\end{align*}


*In (2) we switch back to the cycle notation: $a\to b\to c\to a$.

In the same way we can calculate e.g. the next identity with four elements:
\begin{align*}
\color{blue}{(ad)(ac)(ab)}&
\equiv  \begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\left(\begin{pmatrix}a&b&c&d\\c&b&a&d\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&a&c&d\end{pmatrix}\right)\\
&=\begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&c&a&d\end{pmatrix}\\
&=\begin{pmatrix}a&b&c&d\\b&c&d&a\end{pmatrix}\\
&\,\,\color{blue}{\equiv (abcd)}
\end{align*}
and the claim follows.

