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How can I write these permutations as products of disjoint cycles?

i.$\;\;(1234)(513)$
ii.$\;\;(13526)(53)(46215)$
iii.$\;(13)(12)(32)(143)$

Jordan Glen
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orestiskim
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  • Hello. Can you enclose your math in dollar signs? Such as $(1234)(513)$. It will make it appear nicer. Also, here's a more thorough introduction to how to type in math on this site: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mankind Apr 01 '15 at 12:57

2 Answers2

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I'll spell out how to do it for (i.) and call that a hint for the other two.

For (i.), think about where each element ends up.

$1 \rightarrow 2$ in the first cycle, and $2\rightarrow 2$ in the second cycle. So in the product, $1\rightarrow 2\rightarrow 2$. In short, $1$ goes to $2$.

$2\rightarrow 3$ in the first cycle, and $3\rightarrow 5$ in the second cycle. So in the product $2\rightarrow 3\rightarrow 5$. In short, $2$ goes to $5$.

Since we left at $5$ in the last step, we start at $5$ here. $5\rightarrow 5$ in the first cycle and $5\rightarrow 1$ in the second. In short, $5$ goes to $1$. We've gotten back to where we started, so the first cycle is $(125)$.

Now, we begin again with the first element we haven't seen (in that cycle) which is $3$.

$3\rightarrow 4$ in the first cycle and $4\rightarrow 4$ in the second, so $3$ goes to $4$.

$4\rightarrow 1$ in the first cycle and $1\rightarrow 3$ in the second, so $4$ goes to $3$. This closed the cycle and we have $(34)$. We've now seen all the elements, so we are done.

The disjoint cycle notation is: $(125)(34)$

TravisJ
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  • Thanks a lot!!so for the second the disjoint cycle notation is (1 4 6 5)(2)(3) and for the third it's (1 2 4 3)... – orestiskim Apr 01 '15 at 13:30
  • Yes, that is right. You can suppress the $(2)$ and the $(3)$ (not write them) if you like. Usually, cycles of just one element (fixed points) are not written. – TravisJ Apr 01 '15 at 17:33
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For $(ii)$, we start with $123456 \to 651432 \to 653412 \to 523164$. Identify any fixed points, in this case $2$ and $3$, and this leaves the chain $1\to4\to6\to5\to1$, and so the answer is $(1465)(2)(3)$.

JMP
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