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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!

J. W. Tanner
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daikons
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  • If you were doing $(ac)$ and then $(ab)$, you do not get $(cab)$. You would get: $a\mapsto c\mapsto c$, $b\mapsto b\mapsto a$, and $c\mapsto a\mapsto b$, so you would get $(acb)\neq (cab)$. – Arturo Magidin Nov 18 '20 at 04:37

4 Answers4

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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 $(ac)(ab)=(abc)$.

ndhanson3
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  • How is $(ab)$ fixed by $(ac)$? What exactly does it mean? Also why does $(ab)$ send $a$ to $b$ and also $b$ to $a$? – daikons Nov 18 '20 at 04:41
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    Edited the first bullet to make it clear that $b$ is fixed, not $(ab)$. Also, when we write the permutation $(abcd)$, which is actually called a cycle, we mean $a\mapsto b$, $b\mapsto c$, $c\mapsto d$, and $d$ circles back to the beginning $d\mapsto a$. – ndhanson3 Nov 18 '20 at 04:44
  • Oh I see, so $b \mapsto c$ because $a \mapsto b$? – daikons Nov 18 '20 at 04:48
  • For the original question? Think of $(ac)$ as a function $f$ that swaps a and c and fixes b, and then $(ab)$ as a function $g$ that swaps a and b fixes c. Then $(f\circ g)(b)=f(g(b))=f(a)=c$. This $f\circ g$ represents the product $(ac)(ab)$. – ndhanson3 Nov 18 '20 at 05:00
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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 $(ac)$.

J. W. Tanner
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1

$(a b c) =a\mapsto b, b\mapsto c, c\mapsto a$
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
$a\mapsto b, b\mapsto c, c\mapsto a = (abc)$

novice_2
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1

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.

Markus Scheuer
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