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I have the disjoint cycle: $$(156)(2437).$$ Apparently the "method" would get us: $$(1,6)(1,5)(2,7)(2,3)(2,4).$$ Basically you take the first number, and put it as a transposition of the last number and go backwards till you use all of them. I don't understand where this comes form. I don't know why this method works. Does it even work for my case? Oh and the order is 12 with the permutation being odd.

I think I understand $$(1,5)(5,6)(2,4)(4,3)(3,7)$$ much better. Is this a correct way to do transpositions as well?

H5159
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    Shouldn't it be $(15)(16)(24)(23)(27)$? Permutations are conventionally composed from left to right, and it seems that by comparing the disjoint cycle representation to the product of transpositions representation, that this is where you've had difficulty. – hardmath Apr 02 '14 at 06:34
  • Yeah I don't understand why they follow this pattern. And where I found the method (as an answer to another problem on here), they used the way I ordered it. – H5159 Apr 02 '14 at 06:35
  • @hardmath http://math.stackexchange.com/a/320011/133156 here you go. It seems as if this is a correct answer to, which makes more sense to me $$(1,5)(5,6)(2,4)(4,3)(3,7)$$ Is this correct as well? – H5159 Apr 02 '14 at 06:36
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    I disagree with hardmath in that permutations, being functions, are also often composed from right to left. A bit unnatural, but a consequence of the usual convention of the name of the function preceding that of the variable. Admittedly I have seen it done both ways, but IMVHO (YMMV) it is more common to compose from right to left. – Jyrki Lahtonen Apr 02 '14 at 06:39
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    My book composes from right to left. – H5159 Apr 02 '14 at 06:40
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    @JyrkiLahtonen: it depends on which action you used, left or right. in elemenry books are mostly used left action. – mesel Apr 02 '14 at 06:40
  • I see your point, but I can only guess that @amWhy is composing from right to left in that Question and Answer. – hardmath Apr 02 '14 at 06:41
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    @Frumpy then what is the trouble? Just verify that both expressions are the same permutation, by definition of right-to-left multiplication. – Dan Shved Apr 02 '14 at 06:41
  • I'm having trouble actually computing these because I don't understand how the book computed the permutations in this format. I understand how to "read" these when they are in their disjoint cyclic form but not when they are in their transposition form. – H5159 Apr 02 '14 at 06:43

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Assuming we are composing from right to left, the original product of transpositions:

$$ (1,6)(1,5)(2,7)(2,3)(2,4) $$

can be applied to each item to show it agrees with the product of disjoint cycles:

$$ (156)(2437) $$

For example, let's apply the product of transpositions to $1$. Going from right to left, the first three transpositions leave $1$ fixed as it doesn't appear yet. Then $(1,5)$ sends $1$ to $5$, and since $5$ is not "moved" by the last transposition, that's how $1$ is mapped by the permutation as a whole.

Similarly in the product of disjoint cycles, $1$ is not affected by the first (rightmost) cycle, and in the end we see $1$ goes to $5$ by the second cycle $(156)$.

So these agree as functions on the mapping of $1$ to $5$. Showing the equality amounts to showing they agree on all inputs (elements of the domain).

hardmath
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  • Could you show me what happens when we look at 2, so I can see how it works if the element isn't ignored. Edit: if we look at 2, we see that 2 is sent to 4 by the first transposition, but then 4 does not come up again, so 2 will go to 4? And if we look at 3, we see that 3 is being sent to 2, and then 2 is being sent to 7, which means 3 is being ant to 7? – H5159 Apr 02 '14 at 06:57
  • @Frumpy That's exactly how you do it! – Christoph Apr 02 '14 at 07:05
  • Yes, I think you've got it! In the product of disjoint cycles, it's a little confusing because it reads each cycle from left to right (but since the cycles are disjoint, it doesn't matter what order we put the two cycles in), and $(2437)$ means $2$ goes to $4$, $4$ goes to $3$, $3$ goes to $7$, and finally $7$ goes to $2$. – hardmath Apr 02 '14 at 07:05
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    Yeah that is where my confusion was coming from. It goes left to right and is intuitive in the disjoint form. But in the transposition form, what you do is you start with your first element, find it, going from right to left and see how each transposition you pass changes it. If it doesn't change it you leave it, if it does you go to "that place." So basically I just checked, both the transpositions I have shown are correct. By the way, the permutation is odd because the number of transpositions are odd. And the order is 12 correct? – H5159 Apr 02 '14 at 07:08
  • Yes, if you apply the permutation twelve times, everything gets mapped to itself. Basically, twelve is the least common multiple of the two orders, 3 and 4, of the two disjoint cycles. In the disjoint cycle notation there's no way to tell with certainty what the underlying domain is, but anything that doesn't appear in any of the disjoint cycles stays fixed (so that the permutation is always the identity outside the combined subsets of the cycles). – hardmath Apr 02 '14 at 07:26
  • @hardmath Hey Hardmath, I'm studying for another exam and I have a quick question that is confusing me. If we look at 2, it goes from 2 to 4, and then 4 goes to 3, and 3 goes to 7. But we don't say 2 goes to 7, we say 2 goes to 4. Why? – H5159 Apr 16 '14 at 20:59
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    @Frumpy: You seem to be applying the permutation more than once. If you apply it once, then it maps 2 to 4. Of course the permutation maps 4 somewhere also (it maps all the items 1 through 7 to some value, rearranging them so we call it "permutation"). But if you apply the mapping once, the disjoint cycle representation tells us quite clearly where 2 goes (check your original Question). – hardmath Apr 17 '14 at 00:33
  • Also valid, dear @hardmath, is transposing $(156)(2437)$ in the following manner: $(15)(16)(24)(43)(37)$. (Note that your method and mine are valid when we are working strictly with disjoint cycles.) Hence the original permutation, like you found, is again expressed in an odd number of transpositions, and as the asker stated in their question, they knew the "algorithm" you've used here, but were more comfortable with the algorithm I used. Both are equally correct. There is no unique transposition of a permutation, but only one unique parity, in any given product of disjoint cycle(s). – amWhy Nov 06 '17 at 19:41