0

If $\epsilon = \beta_1\beta_2 \cdots \beta_r$ , where the $b$’s are $2$-cycles, then $r$ is even.

The following is part of a proof of the above theorem.

proof.

Clearly $r\neq1$, since a $2$-cycle is not the identity. If $r=2$, we are done. So, we suppose that $r>2$, and we proceed by induction. Suppose that the rightmost $2$-cycle is $(ab)$. Then, since $(ij) = (ji)$, the product $\beta_{r-1}\beta_r$ can be expressed in one of the following forms:

$\epsilon = (ab)(ab)$,

$(ab)(bc) = (ac)(ab)$,

$(ac)(cb) = (bc)(ab)$,

$(ab)(cd)=(cd)(ab)$.

How do we get the second form,$(ab)(bc) = (ac)(ab)$, given that $(ij)=(ji)$?

Community
  • 1
Skm
  • 2,392
  • 3
    Both equal $(a,b,c)$. – Angina Seng Oct 25 '17 at 19:26
  • 1
    There are only three symbols... you can just look at where they all map for both maps. – rschwieb Oct 25 '17 at 19:26
  • Why not track the image of each element in ${a,b,c}$ by $(ab)(bc)$ and by $(ac)(ab)$, and compare them? – Did Oct 25 '17 at 19:29
  • Comparing them, I see why they both are essentially equivalent to $(a b c)$, but I thought that there was a way to prove it without using direct computation. – Skm Oct 25 '17 at 19:35
  • My guess is that the proof cites the fact that $(ij) = (ji)$ in order to explain why $(ca)(ab)$ and $(cb)(ab)$ are not listed as separate cases from $(ac)(ab)$ and $(bc)(ab).$ Otherwise it seems pointless to mention that fact. – David K Oct 25 '17 at 19:49
  • @DavidK: That makes sense. – Skm Oct 25 '17 at 19:50
  • You might also want to take a look at this: https://math.stackexchange.com/q/46403/9464 –  Oct 25 '17 at 19:53

0 Answers0