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I just want to address the idea of disjoint cycles and product of cycles because I believe I am confusing the idea.

My professor said:

Write $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ as a product of disjoint cycles in $S_7$

where the answer is: $$(2\:5)(6\:4\:7)(2\:4\:5\:3)=(2\:7\:6\:4)(3\:5)$$

but the idea here is he moved from right to left.

Now I'm working out problems and I encountered this problem:

Consider the permutation $a=(1\:4\:6\:3\:7)(2\:8)$ and $b=(2\:5\:3)(4\:7\:8\:1)$ in $S_8$

now calculate:

$$ab=(1\:4\:6\:3\:7)(2\:8)(2\:5\:3)(4\:7\:8\:1)=(1\:7\:4\:6\:2)(3\:8\:5)$$

but in order to calculate the product, the work was done from left to right?

So my question is why for disjoint cycles, you move from right to left but for products left to right?

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    $\begingroup$ I'm confused, $(14637)\in S_5$? or$S_8$? $\endgroup$ – user217174 Mar 12 '15 at 20:24
  • $\begingroup$ Sorry, $S_8$ made a typo. $\endgroup$ – mika Mar 12 '15 at 20:28
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    $\begingroup$ It's sloppy to move from right to left in one problem and left to right in another. There's no universal convention though. So pick one and stick to it. If your professor does both, then I'd talk to him/her, complain, or something of the like. It's simply bad style. Same goes with function composition...do it one way and stick with it. $\endgroup$ – Daniel W. Farlow Mar 12 '15 at 20:41
  • $\begingroup$ @Crash, alright! Because I'm use to right to left since that's normally how I was taught to do compositions, so the left to right threw me off a bit. and one of the person who answer mention disjoint cycles are commutative so I will stick with one way then. Thanks! $\endgroup$ – mika Mar 12 '15 at 21:04
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For the first one, writing $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ as a product of disjoint cycles means that you must write it in a way no number is repeated in any other cycle. In the cycles $(2\:5)$ and $(2\:4\:5\:3)$ we clearly see that $2$ and $5$ are repeated. So what can we do? Firstly write this product as a cycle. Take any number, for example $2$. We have $(2\:5)(6\:4\:7)(2\:4\:5\:3)$. Thus the image of 2 according to $(2\:4\:5\:3)$ is 4 and the image of 4 according to $(6\:4\:7)$ is 7 and the image of 7 according to $(2\:5)$ is itself,i.e, 7 since it doesn't appear in the cycle. So the image of 2 according to $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ is 7. Start writing $(2\:7$. Now what's the image of 7? Do the same think, you'll get 6. $(2\:7\:6$. The image of 6 is 4. $(2\:7\:6\:4$. The image of 4 is 2. $(2\:7\:6\:4\:2$. WAIT A MINUTE! 2 again! So we stop here: $(2\:7\:6\:4)$. But in $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ there are numbers not in $(2\:7\:6\:4)$ which are 3 and 5. According to $(2\:5)(6\:4\:7)(2\:4\:5\:3)$, the image of 3 is 5 and the image of 5 is 3, $(3\:5)$. We have all the numbers here: $(2\:7\:6\:4)$ and $(3\:5)$. Thus:$$(2\:5)(6\:4\:7)(2\:4\:5\:3)=(3\:5)(2\:7\:6\:4)=(2\:7\:6\:4)(3\:5)$$

We have commutivity because the cycles are disjoint.

Edit: For your question left or right, it's comes from commutativity of disjoint cycles:$$a=(1\:4\:6\:3\:7)(2\:8)=(2\:8)(1\:4\:6\:3\:7)$$

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  • $\begingroup$ Thanks a lot. I understand now. $\endgroup$ – mika Mar 12 '15 at 21:04
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    $\begingroup$ I'm glad to hear it. You're welcome! $\endgroup$ – Scientifica Mar 12 '15 at 22:38

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