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I'm trying to find the order of $\sigma = (1\ 2\ 3\ 5)(4\ 5\ 6\ 7)$ by first composing it into disjoint cycles. I thought that 1 must go to 2, 2 then goes to 3, 3 goes to 5, 5 goes to 6, 6 goes to 7 and 7 to 4, however this doesn't give me disjoint cycles. (Or one which makes sense). I use the notation of working from right to left, but if I want to start with 1 then it seems that I actually start is the most left cycle. Can someone please explain where my work went wrong? Also, how can you figure out what would the order be if you end up with with one long cycle? Thank you!

Layla16
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  • Try writing it in two-line form and then finding the disjoint cycles from that. – JMoravitz Apr 18 '24 at 17:52
  • I tried this, but then got stuck as I wasn't sure what would map to 5, i.e. if either 3 mapped to 5 or 4 mapped to 5 – Layla16 Apr 18 '24 at 17:53
  • You wrote "3 goes to 5" but this is wrong. $3$ will go to $5$ if we only had the left part, but then the result of that is still affected by the right part. Here we have $3\mapsto 5\mapsto 6$ and so simplified we have $3$ goes to $6$. The correct two-line for this would be $\begin{pmatrix}1&2&3&4&5&6&7\2&3&6&5&1&7&4\end{pmatrix}$ – JMoravitz Apr 18 '24 at 17:54
  • We can see then that this does simplify to just one long cycle. The order of a cycle is the length of the cycle. The order of multiple disjoint cycles is the LCM of the lengths of those cycles. – JMoravitz Apr 18 '24 at 17:55
  • Thank you for your comments, can you perhaps explain a bit further why 3 actually goes to 6? I thought (despite this not being very formal) that if you "go-to" a number which appears in other cycles, that you only look to further cycles ( ones to the left of the one you're in) to see what number to actually go to, and not past ones. Is this not the case? – Layla16 Apr 18 '24 at 17:59
  • Is there perhaps also an easy way to visualise what this would look like in two-line form? – Layla16 Apr 18 '24 at 18:00
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    Ah, perhaps you multiply in the opposite order than I am used to. That's fine then... we would have $4\mapsto 5\mapsto 1$ and would have then $\begin{pmatrix}1&2&3&4&5&6&7\2&3&5&1&6&7&4\end{pmatrix}$. The punchline is that one of these elements gets moved twice in a single application of $\sigma$, once from the cycle $(1~2~3~5)$ and once from the cycle $(4~5~6~7)$. Which one acts on the elements first varies based on the convention you use, but the general behavior here is the same. – JMoravitz Apr 18 '24 at 18:00
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2 Answers2

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You correctly calculated $$\sigma = (1\ 2\ 3\ 5)(4\ 5\ 6\ 7)=(1\ 2\ 3\ 5\ 6\ 7\ 4).$$ The order of a cycle (here: $\sigma$) is it length (here: $7$).

Anne Bauval
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Hint: Remember that

$$\sigma=(1235)(4567)$$

is the composition of two functions:

$$\begin{align} \sigma_1(x)&:=(1235)(x)\\ &=\begin{cases} 1 &: x=5\\ 2 &: x=1\\ 3 &: x=2\\ 5 &: x=3\\ x &: o/w \end{cases} \end{align}$$

and

$$\begin{align} \sigma_2(x)&:=(4567)(x)\\ &=\begin{cases} 4 &: x=7\\ 5 &: x=4\\ 6 &: x=5\\ 7 &: x=6\\ x &: o/w. \end{cases} \end{align}$$

The order of a permutation is the least common multiple of the orders of its disjoint cycles. The order of a cycle is its length.

Shaun
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