6
$\begingroup$

When do two permutations commute?

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

$\endgroup$
  • 6
    $\begingroup$ Conjugate permutations hardly ever commute. $\endgroup$ – user228113 Nov 13 '17 at 23:50
  • $\begingroup$ $ ab = ba \iff b = a^{-1}ba $ ? $\endgroup$ – Zaz Nov 13 '17 at 23:52
  • $\begingroup$ 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. $\endgroup$ – user228113 Nov 13 '17 at 23:54
  • $\begingroup$ @G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then? $\endgroup$ – Zaz Nov 13 '17 at 23:58
  • 3
    $\begingroup$ The question is a duplicate of this. $\endgroup$ – Alex Ravsky Nov 23 '17 at 3:45
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ In support of this: $P_{12}P_{23}\ne P_{23}P_{12}$ although both elements have the same cycle structure. $\endgroup$ – user160660 Nov 13 '17 at 23:57
2
+50
$\begingroup$

The answer below needs an edit, please see discussion below.


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.

$\endgroup$
  • $\begingroup$ is this a necessary and sufficient condition for two permutation to commute? Thank you. $\endgroup$ – GA316 Dec 17 '18 at 10:03
  • 1
    $\begingroup$ 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 $\pi\sigma\pi^{-1}=\sigma$. $\endgroup$ – Alexander Burstein Dec 17 '18 at 15:25
  • $\begingroup$ Thank a lot for the clarification :) $\endgroup$ – GA316 Dec 17 '18 at 16:19
  • $\begingroup$ 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. $\endgroup$ – GA316 Dec 19 '18 at 10:35
  • $\begingroup$ 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. $\endgroup$ – Alexander Burstein Dec 19 '18 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.