Write (15)(286)(479) in 1-line notation. Find the inverse of (15)(286)(479).
Can anyone please help?
Thank you.
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1Am I missing something or is your permutation already in one line notation? – John Habert Feb 12 '14 at 17:35
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Hi @Lienhua. Welcome! What have you tried so far? (Also: you might find this tutorial helpful $\ddot\smile$) – Shaun Feb 12 '14 at 17:37
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The problem is stated like that. I would say it looks like it is in 1-line notation. – user1523 Feb 12 '14 at 17:44
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For the second, I have tried writing it in a reverse order. For example, (974)(682)(51). I am not sure if this is the correct answer. – user1523 Feb 12 '14 at 17:46
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@JohnHabert: I think what is given is in cycle notation. One-line notation is like the ordinary two-line permutation notation, you just omit the implicit upper line since it is just "1 2 3 ... n". – MPW Feb 12 '14 at 18:27
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@MPW Thanks for the info. Considering how prevalent cycle notation is, one-line notation written in the what looks exactly like cycle notation seems like it would just lead to confusion. – John Habert Feb 12 '14 at 18:32
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@JohnHabert: I agree completely. I've never seen anybody use 1-line notation. Seems a little silly, doesn't it -- the presence of 2 lines is the only thing that distinguishes it from cycle notation. – MPW Feb 12 '14 at 18:35
3 Answers
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Hint Disjoint cycles commute. So you can test whether your inverse is actually an inverse by just checking what happens when you multiply $(479)(974)$ and $(286)(682)$. A 2-cycle like $(15)$ is always its own inverse.
John Habert
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So, for example (479)(974) = (9)(7)(4) ? What can I conclude from this result? – user1523 Feb 12 '14 at 17:58
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$(9)$ by itself means that the permutation doesn't do anything to $9$. In other words, you've gotten back the identity. – John Habert Feb 12 '14 at 18:00
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One-line notation is just the ordered list of images of $1,2,\ldots,n$. So it would be $(5\;8\;3\;7\;1\;2\;9\;6\;4)$, right?
Then the inverse would be $(5\;6\;3\;9\;1\;8\;4\;2\;7)$ by observation.
MPW
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Let me consider a similar permutation: $$\sigma:=(1\:8\:3)(2\:9)(4\:6\:7).$$ Now, since $9$ is the greatest number that appears in the disjoint cycles, then we can assume that $\sigma$ is a permutation on $\{1,2,3,4,5,6,7,8,9\}.$ In particular, $\sigma$ takes $1\mapsto 8\mapsto 3\mapsto 1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto6\mapsto7\mapsto4,$ and (by process of elimination) $5\mapsto 5.$ The inverse of $\sigma$ will undo all of these mappings--meaning it needs to send $1\mapsto 3\mapsto 8\mapsto1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto 7\mapsto 6\mapsto 4,$ and $5\mapsto 5$. Put back into disjoint cycle notation, $$\sigma^{-1}=(1\:3\:8)(2\:9)(4\:7\:6),$$ or equivalently, $$\sigma^{-1}=(1\:8\:3)^{-1}(2\:9)^{-1}(4\:6\:7)^{-1}.$$ There are other ways we could show this, of course, since $(1\:3\:8)=(3\:8\:1)=(8\:1\:3),$ for example, but this does the trick.
As for rewriting $\sigma$ in one-line notation, it ultimately amounts to figuring out where $\sigma$ takes $1,2,3,4,5,6,7,8,$ and $9,$ respectively. Since $\sigma(1)=8,$ $\sigma(2)=9,$ $\sigma(3)=1,$ $\sigma(4)=6,$ $\sigma(5)=5,$ $\sigma(6)=7,$ $\sigma(7)=4,$ $\sigma(8)=3,$ and $\sigma(9)=2,$ then the one-line notation is: $$8,9,1,6,5,7,4,3,2$$ This is simply a condensed form of the two-line notation: $$\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\8 & 9 & 1 & 6 & 5 & 7 & 4 & 3 & 2\end{pmatrix}$$
Cameron Buie
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