Which of the following numbers can be orders of a permutation $\sigma$ of $11$ symbols, such that $\sigma$ does not fix any symbols?
$1. \;18$
$2.\; 30$
$3.\;15$
$4.\; 28$
could any one just give me hints?
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What do you know about the order of a permutation? – Thomas Andrews May 31 '13 at 16:53
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length of the permutation is its order. – Myshkin May 31 '13 at 16:56
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1What is the order of $(1 2)(3 4 5)$? – Mark Bennet May 31 '13 at 16:57
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3$6$, in this case I need to take lcm of order of the cycle – Myshkin May 31 '13 at 16:58
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1OK so can you get a cycle pattern on 11 symbols with lcm equal to the various target values? – Mark Bennet May 31 '13 at 17:05
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1I can get a permutation which is product of a $5$ cycle and a $6$ cycle hence $30$ will be an answer? similarly a $7$ cycle and a $4$ cycle hence $28$ will be another answer? and a $9$ cycle and a transposition. so $18$ will be another answer – Myshkin May 31 '13 at 17:07
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Looks good to me – Mark Bennet May 31 '13 at 17:09
2 Answers
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Think about the orders of the permutations (1 2) (3 4 5) and (1 2) (3 4 5 6) and see if you can think of a rule that gives the order of a permutation written in disjoint cycle notation. Now if $\sigma$ fixes no points, what does it tell you about the permutation written in disjoint cycle notation? From this you should be able to tell which orders are possible.
john
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@TaxiDriver - you perhaps need to look more carefully at 15 - hint you can have as many cycles as you like. – Mark Bennet May 31 '13 at 17:13
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Close, but not quite. Can you give permutations that have these orders? Also can you say why order 15 is impossible? – john May 31 '13 at 17:14
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well, I am not able to find an element of $S_{11}$ whose order is $15$ – Myshkin May 31 '13 at 17:19
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2There are definitely elements of order $15$ in $S_{11}$, but finding one which fixes no points is, perhaps, a little harder. – Thomas Andrews May 31 '13 at 17:31
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any $5,6$ cycle(disjoint), any $9,2$ cycle(disjoint), any $7,4$ cycle(disjoint). – Myshkin May 31 '13 at 18:58
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I am not able to see why $15$ will be also an answer, it would be great if some one tell me clearly, have exam very near :) – Myshkin May 31 '13 at 19:02
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2an element of order 15 would have to have either 15-cycles or 5-cycles and 3-cycles (or some combination). Now it's just a matter of finding a combination of such cycles whose lengths sum to 11. 11=5+3+3 so we could have (1 2 3 4 5)(6 7 8)(9 10 11) as an example – john Jun 01 '13 at 00:58
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Firstly find all partitions of 11 as follows: $$\begin{array}{ll} & \text{ corresponding l.c.m } \\ 11 &11\\ 10+1. &10 (\text{l.cm of }10,1)\\ 9+2. & 18 \text{l.cm of }9,2 \\ 9+1+1\\ 8+3\\ 8+2+1\\ 8+1+1+1\\ .\\ .7+4\\ .\\ .\\ 6+5\\ 6+4+1\\ .\\ .\\ 5+5+1\\ .....\\ 5+ 3+2\\ 5+3+1+1+1. & 15( \text{l.c.m of } 5,3,1,1)\\ .......\\ \text{Similarly up to } 1+1.....+1 \ (11 \text{times})\\ \end{array} $$ When we calculate l.c.m in all above cases we see 28 l.c.m is not possible So possible answers are 18,15,30
Calvin Khor
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1Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Nov 02 '19 at 12:31