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Disjoint cycles commute: $(ab)(cd) = (cd)(ab)$, but do non-disjoint cycles commute? Does $(ac)(ab) = (ab)(ac)?$

Consider the composition of two permutations: $\begin{pmatrix} a & c\\ c & a \\ \end{pmatrix}$$\begin{pmatrix} a & b\\ b & a \\ \end{pmatrix}.$

Right to left: $f(a) = b$, but there's no function(?) $g$ that sends $b$ to another value. How do we deal with it?

user255751
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  • Cycles that are not disjoint don't usually commute. Consider $(12)(13)$ and $(13)(12)$. The first sends $1$ to $3$, the second sends $1$ to $2$. – Pedro Jul 21 '15 at 18:04
  • You should mention your set on which you taking permutation – Chiranjeev_Kumar Jul 21 '15 at 18:04
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    You're writing your permutations as 'matrices' in an unusual way. If I had to write the cycle $(ab)$ using the matrix-notation, I would use $\begin{pmatrix}a&b&c&d\b&a&c&d\end{pmatrix}$. This makes it clear that $c$ and $d$ are acted on, they just don't go anywhere. – pjs36 Jul 21 '15 at 18:05

2 Answers2

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The two permutations can be written as $$ g = \pmatrix{a&b&c\\c&b&a}, \quad f = \pmatrix{a&b&c\\b&a&c} $$ We have $$ g(f(a)) = g(b) = b\\ f(g(b)) = f(c) = c $$

Ben Grossmann
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The cycle $(a_1\,a_2\,\ldots\, a_n)$ dentoes a permutation that maps $a_1\mapsto a_2$, $a_2\mapsto a_3$, ..., $a_{n-1}\mapsto a_n$, $a_n\mapsto a_1$ and for any other element it leaves it untouched, i.e., it maps $x\mapsto x$.

Therefore, $(a\,c)(a\,b)$ first maps $a\mapsto b$, then leaves $b$ untouched, i.e., in total it maps $a\mapsto b$. It also maps $b\mapsto a$ and then $a\mapsto c$, so in summary $b\mapsto c$; and first $c$ is untouched and then $c\mapsto a$; finally anything else is left untouched. You may notice that therefore $(a\,c)(a\,b)=(a\, b\,c)$. (On the other hand, $(a\,b)(a\,c)=(a\,c\,b)$).

Hagen von Eitzen
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  • Let me try: $(ab)(ac)$ is actually $[(ab)(c)][(ac)(b)]$. So, $ (ac)(b)$ means $a \mapsto c.$ Then, $(ab)(c)$ means $c \mapsto c.$ So far we have $(a c \ldots) $. Further, $(ac)(b)$ means $c \mapsto a$ and $(ab)(c)$ means $a \mapsto b.$ In all, $(acb).$ Do you approve? – user255751 Jul 21 '15 at 19:14