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!
 A: These cycles are effectively functions (think $f\circ g$). Since you are asking what the composition of functions $(ac)(ab)$ does to the elements $a$, $b$, and $c$, you need to read right to left, just as you would for $f(g(x))$.
$\bullet$ $(ab)$ sends $a$ to $b$. Then $(ac)$ fixes $b$. So we get $a\mapsto b$.
$\bullet$ $(ab)$ sends $b$ to $a$. Then $(ac)$ sends $a$ to $c$. So we get $b\mapsto c$.
$\bullet$ $(ab)$ fixed $c$. Then $(ac)$ sends $c$ to $a$. So we get $c\mapsto a$.
So we have $(ac)(ab)=(abc)$.
A: There are different conventions, but here we are performing the permutation on the right first.
So $a\mapsto b$ by $(ab)$, and the result $b$ is unaffected by $(ac)$;
$b\mapsto a\mapsto c$ by $(ab)$ followed by $(ac)$;
and $c$ is unaffected by $(ab)$ but $c\mapsto a$ by $(ac)$.
A: $(a b c) =a\mapsto b, b\mapsto c, c\mapsto a$
For $(a c)(a b) $ we need to start the operation from rightmost, so 
$(a b)  = a\mapsto b, b\mapsto a, c\mapsto c$, when we apply $(a c)$ on this we get
$a\mapsto b\mapsto b, b\mapsto a\mapsto c, c\mapsto c\mapsto a$, so the composite map is
$a\mapsto b, b\mapsto c, c\mapsto  a = (abc)$
A: 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.

