(a) $(1,2,4)(4,3,5)(2,4)(1,2,4)^{-1}$

In the solution for this question, my professor has the product of disjoint cycles written as (1,3,5)(4,1). How would this make sense when disjoint cycles are not allowed to have repeat elements within cycles, in this case, the number 1?

I ended up with $P=\begin{bmatrix}1&2&3&4&5&6 \\ 4&2&5&3&1&6\\ \end{bmatrix}$ which can be expressed as one cycle = (5 1 4 3)

(b) (1,3,4,5,2)(1,2)(1,3,5)(2,5,4,3,1)

For part (b), he first inversed the last permutation so this becomes $(1,3,4,5,2)(1,2)(1,3,5)(1,3,4,5,2)^{-1}$ then wrote the product of disjoint cycles as (3,1)(3,4,2). Is there a reason why he inversed the last permutation (2,5,4,3,1) into $(1,3,4,5,2)^{-1}$ before calculating the product?

I would greatly appreciate any help with understanding the reasoning behind his answers and how to deal with inverse permutations when performing these operations.


The idea is that conjugation by an element, i.e. $(124)g(124)^{-1}$, can be computed by switching the elements of $g$ as specified by the element you are conjugating by, which is $(124)$.

So we should take $g =(435)(24)$ and to conjugate by $(124)$, switch 1s to 2s, 2s to 4s, and 4s to 1s. This gives $(135)(41)$. You are correct that this is not a product of disjoint cycles, the disjoint cycle form would be $(1435)$.

The second example is the same idea. You are correct also that $(124)^{-1} = (421)$.

  • $\begingroup$ Got it, thanks a lot for the answer. I appreciate your help. May I know if there's a theorem with the conjugating property you mentioned above or did you derive that logically? $\endgroup$ – Squires McGee Oct 18 '14 at 4:11
  • $\begingroup$ A theorem is not necessary. I'm not sure how to explain properly, other than to say it is exactly like doing a change of basis for a matrix. Consider $(124)(23)(421)$. Doing $(421)$ first is kind of like relabeling the set $\{1,2,3,4,5,6\}$ so that $(23)$ would see 4 as a 2 and switch 3 and 4 instead, i.e.$ (23)(421) = (1432)$. Then doing $(124)$ at the end switches things back to normal labeling, and all that has happened is we have swapped 3 with 4. So any conjugated element does just what it normally does, but thinks that 2s are 1s, 4s are 2s and 1s are 4s. $\endgroup$ – Ben Oct 20 '14 at 6:59

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