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please explain I am trying to learn abstract algebra. I would appreciate any abstract algebra recourses you have to offer in general. Links to books, videos etc.

Jean Marie
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Taylor
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  • Did you try to search yourself already? This site also has a lot of resources, e.g.,here. – Dietrich Burde Feb 21 '19 at 20:37
  • yes I have. Sorry maybe shouldn't have made a reference request and just asked the question (which Im having trouble understanding). ty for the link – Taylor Feb 21 '19 at 20:46
  • Im not exactly sure what you mean. Just getting into this and having a lot of trouble with the terminology and such. I struggle with new math classes at first and then it all starts to come together. Trying to find some really clearly worked examples. – Taylor Feb 21 '19 at 20:54
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    I have taken the liberty to suppress the "PLEASE EXPLAIN" in capitals that has nothing to do in a title. You do not need to "trumpet" the fact that you desire explanations... – Jean Marie Feb 21 '19 at 21:11
  • If that were true the only possible orders for the elements of $S_{60}$ would be $1,3,4,12$. On the other hand $S_{60}$ is full of elements with order $p\in{2,5,7,11,13,\ldots}$ by Cauchy's theorem. – Jack D'Aurizio Feb 21 '19 at 21:12
  • Sorry Jean and thank you. Im new here – Taylor Feb 21 '19 at 21:15

1 Answers1

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Every element of $S_n$ has a unique decomposition into the composition of disjoint cycles. Consider $(1,2,3,\dots,60)\in S_{60}$. This is already a composition of disjoint cycles, so it is unique. But it does not have length $3$ or $4$! What can you conclude?

YiFan Tey
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  • Does the OP understand notation $(1,2,3,\cdots,60)$ which means that the image of $1$ is the element $2$ at its right, that the image of $2$ is $3$ for the same reason... and finally that the image of $60$ is $1$ ? – Jean Marie Feb 21 '19 at 21:07
  • Thank you both! I do have this understanding but am not always sure of myself so having every bit explained really helps. Would the answer simply be to break this up into cycles of length 3 and 4 $\in S_{60}$ – Taylor Feb 21 '19 at 21:14
  • @Taylor The point is that whenever an element of $S_{60}$ has been broken up into disjoint cycles, you can't break it up further, or express it in any other way except by reordering the cycles (hence "unique"). The element $(1,2,\dots,60)$ is already in disjoint cycle decomposition, so we can't break it up further. We have found a counterexample to the claim that every element of $S_{60}$ is the disjoint composition of $3$ and $4$-cycles. So the answer to the question in your title is it is false. – YiFan Tey Feb 21 '19 at 21:16
  • @Dietrich Burde I share your opinion – Jean Marie Feb 21 '19 at 21:19
  • This helps and yes I will work through more basic notation. Thank you for answering – Taylor Feb 21 '19 at 21:24
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    @YiFan I now understand. $S_{60}$ is the group of permutations on a set of 60 elements. (1,2,...,60) is one of the elements and is not a composition of disjoint cyclic permutations of lengths 3 and 4. It is length 60 and cannot be expressed any other way. Am I missing anything? – Taylor Feb 21 '19 at 23:02
  • @YiFan I realize I used the word elements in two different ways in my last comment. Is one of these usages incorrect? – Taylor Feb 21 '19 at 23:03
  • @Taylor if you want to be more unambiguous you can say "$S_n$ is the group of permutations on $n$ letters" or "on a set with $n$ elements", but your usage is fine. And yes, your understanding of the proof is right. – YiFan Tey Feb 22 '19 at 01:39