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.

  • $\begingroup$ 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

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

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

  • $\begingroup$ 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

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