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When do two permutations commute?

How do you find out something like how many permutations in $S_7$ commute with $(12)(345)$?

Zaz
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    Conjugate permutations hardly ever commute. –  Nov 13 '17 at 23:50
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    $ ab = ba \iff b = a^{-1}ba $ ? – Zaz Nov 13 '17 at 23:52
  • I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself. –  Nov 13 '17 at 23:54
  • @G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then? – Zaz Nov 13 '17 at 23:58
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    The question is a duplicate of this. – Alex Ravsky Nov 23 '17 at 03:45
  • Here is a hint towards the answer. Given a cycle $\sigma=(i_1\ i_2\dots i_k)$, its conjugate $\pi\sigma\pi^{-1}=(\pi(i_1)\ \pi(i_2)\dots \pi(i_k))$. In other words, conjugation by $\pi$ permutes the positions and values of $\alpha$ according to $\pi$. – Alexander Burstein Nov 26 '17 at 23:30

2 Answers2

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Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.

Robert Israel
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    In support of this: $P_{12}P_{23}\ne P_{23}P_{12}$ although both elements have the same cycle structure. –  Nov 13 '17 at 23:57
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The answer below is incorrect as stated and needs an edit, please see discussion below. And please don't upvote this, it's bad enough that this answer has gotten a bounty.

In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $\sigma$ and $\pi$ commute when $\pi$

  • permutes elements within disjoint cycles of $\sigma$, and/or
  • permutes the sets of elements in equal length disjoint cycles of $\sigma$.

Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set $\{1,2\}$ onto set $\{3,4\}$ and vice versa.

Alexander Burstein
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  • is this a necessary and sufficient condition for two permutation to commute? Thank you. – GA316 Dec 17 '18 at 10:03
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    It’s easier to think of the conditions above as those for $\pi\sigma\pi^{-1}=\sigma$. Also $\sigma: i\mapsto j$ if and only if $\pi\sigma\pi^{-1}: \pi(i)\mapsto\pi(j)$. So, yes, the above conditions are necessary and sufficient for $\pi\sigma\pi^{-1}=\sigma$. – Alexander Burstein Dec 17 '18 at 15:25
  • Thank a lot for the clarification :) – GA316 Dec 17 '18 at 16:19
  • I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $\sigma$"?. Thank a lot. – GA316 Dec 19 '18 at 10:35
  • You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise. – Alexander Burstein Dec 19 '18 at 15:30
  • Thanks. Also can you please made it precise, that would be very helpful for later visitors like me :) . – GA316 Dec 20 '18 at 10:34