0
$\begingroup$

How to show that cycles of even degree are odd permutations? What I understand is that cycles are actually referring to cyclic groups (if not, please correct me). So we have some cyclic group $\langle g \rangle$ such that $|\langle g \rangle | = n\;mod\;2 = 0$. If given a set $\Omega$ with $g\in G\in Sym(\Omega)$, show that $\langle g \rangle$ can be written as a product of an odd number of transpositions.

$\endgroup$
  • $\begingroup$ How about we say cycles of even length rather than even "degree"? This accords with terminology for permutations. $\endgroup$ – hardmath Oct 1 '15 at 0:55
  • $\begingroup$ @hardmath, be my guest, but I just used the exact terminology from my book. $\endgroup$ – bourbaki4481472 Oct 1 '15 at 0:59
  • $\begingroup$ Fair enough, but I'm wondering why you do not rely on the definition your book uses for a cycle permutation and the classification of cycles as even or odd degree? $\endgroup$ – hardmath Oct 1 '15 at 1:06
  • $\begingroup$ @hardmath, because this is from an introduction section of a short book so the author didn't go through all the definitions carefully. $\endgroup$ – bourbaki4481472 Oct 1 '15 at 2:34
2
$\begingroup$

$(1\ 2\ \dots n) = (1\ n)(1\ n-1)\cdots(1\ 2)$

This shows an $n$-cycle can be written as (a product of) $n-1$ transpositions (just replace $k$ with $a_k$).

$\endgroup$
  • $\begingroup$ You must be a right-to-left composer. Many times the composition of permutations will be (by convention) done left-to-right, contrary to the direction we typically interpret function composition. $\endgroup$ – hardmath Oct 1 '15 at 1:10
  • $\begingroup$ @hardmath thanks for mentioning right-to-left and left-to-right - I would have been confused for a little bit. $\endgroup$ – bourbaki4481472 Oct 1 '15 at 2:33
  • $\begingroup$ @hardmath...indeed (because I think of permutations as functions). I am aware of the other convention, and no doubt should have added a note. Too late now :P $\endgroup$ – David Wheeler Oct 1 '15 at 11:15

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.