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I'm working on this question:

List the permutations of $\{1,2,3\}$ in disjoint cycle form.

I already know what a disjoint cycle is. It's basically means that every cycle contains numbers that are not in any other cycle. So with that in mind, do I write all the possible permutations of

\begin{pmatrix} 1 & 2 & 3\\ ? & ? & ? \end{pmatrix}

such that I could write down all of it's permutations in disjoint cycle form? A push to get me started is all I'm asking for here.

EDIT Please see finalized answer below.

John W. Smith
  • 1,597
  • I know that I can arrange the $2nd$ row $3!$ (or $6$) ways. – John W. Smith Mar 30 '17 at 02:50
  • If the bottom row is $3\ 2\ 1$, can you write the resulting permutation using cycle notation? – pjs36 Mar 30 '17 at 02:55
  • $(1 3)(2)$ correct? – John W. Smith Mar 30 '17 at 02:56
  • Right! And so you've found one of the six permutations, written in disjoint cycle form. (It's common also to just write $(1\ 3)$, ignoring cycles with a single element, but neither is "the right way") – pjs36 Mar 30 '17 at 02:59
  • Ahh so once I collect all of them, I will have some repeats. So $(1 3) = (3 1)$ correct? And that will count as $1$ instead of $2$ permutations. – John W. Smith Mar 30 '17 at 03:00
  • Yeah, representations for disjoint cycles aren't unique, $(1\ 3) = (3\ 1)$, or even $(1\ 2)(3\ 4\ 5) = (4\ 5\ 3)(2\ 1)$. So for each of the $3!$ ways to arrange the bottom row, you'll have some freedom in how you write the permutation in cycle notation (just choose one), but distinct permutations have legitimately distinct cycle notations. – pjs36 Mar 30 '17 at 03:05
  • Ahh alright. In that case, I will post a full solution and we'll see if it comes our correct. – John W. Smith Mar 30 '17 at 03:10

1 Answers1

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Here is my attempted:

\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} is given by $(1 3)(2)$. Then \begin{pmatrix} 1 & 2 & 3\\3 & 1 & 2 \end{pmatrix}is given by $(1 3 2)$. Then \begin{pmatrix} 1 & 2 & 3\\ 2& 3 & 1\end{pmatrix} is given by $(1 2 3)$. Then \begin{pmatrix} 1 & 2 & 3 \\2 & 1 & 3 \end{pmatrix} is given by $(1 2)(3)$. Then \begin{pmatrix} 1 & 2 & 3 \\1 & 3 & 2 \end{pmatrix} is given by $(1)(2 3)$.

Lastly, \begin{pmatrix} 1 & 2 & 3\\1 & 2 & 3 \end{pmatrix} is given by $(1)(2)(3)$, but these are just the identity matrices, so we can eliminate $(2)$ and $(3)$ and are only left with $(1)$.

So therefore, the final list of permutations is given by $$(1),(12),(13),(23),(123),(132).$$

John W. Smith
  • 1,597