Number of Elements of Order 5 in S7 Permutation Group

In summary, the number of elements of order 5 in the permutation group S7 is 504. This is calculated by taking into consideration the l.c.m of the lengths of disjoint cycles, which in this case must be 5. The only possible way to achieve this is by having a 5-cycle and a 2-cycle. However, since the l.c.m would then become 10, this is not a valid consideration. Therefore, the correct solution is to have only 5-cycles, giving a total of 504 elements of order 5 in S7.
  • #1
astronut24
10
0
what is the number of elements of order 5 in the permutaion group S7??
so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7 to themselves...so we end up getting 2 cycles of length 1 each. the remaining five elements can be arranged in 4! ways...
so, the answer is 7C2 * 4! = 21*24 = 504.
:smile: but unfortunately, this answer is WRONG!
CAN ANYBODY TELL ME WHY?! PLEEEEASE HELP!
 
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  • #2
Since 5 is prime, you MUST have cycle of length 5.
Then there could be a 2-cycle in addition.
 
  • #3
hello

well...yes, it's possible to have a cycle of length 2 in addition to the 5 cycle...but then the l.c.m becomes 10. so that rules out such a consideration!
 
  • #4
Yes, so the only permutations in S7 of order 5 are the 5-cycles.
 
  • #5
Your answer is wrong because you've counted, for example:

12345, 23451, 34512, 45123, and 51234 as different elements.
 
  • #6
ya...so what further? that's a valid point you've raised...
so do you divide by 4?
 
  • #7
You have [itex]{7 \choose 5}[/itex] ways to pick 5 elements from a set of 7.
There are 5! ways you can order 5 elements in a cycle.
For a given cycle of length 5, 5 orderings are give the same permutation.
 
  • #8
hey galileo

please read stuff carefully...
we left this a long while ago, right astronut?! :smile:
 
  • #9
i think the solution 504 is correct...
i don't see any fallacy in it.
 
  • #10
The general solution is:

suppose t is a permutation of type

[tex]1^{m_1}2^{m_2}...r^{m_r}[/tex]

then the order of the centralizer is

[tex]\prod_i i^{m_i}m_i![/tex]

in this case it is 1^2.5

so the centralizer's order is

2!.5

hence the conjugacy class has order

7!/10

which is indeed 504
 

1. What is the "Number of Elements of Order 5 in S7 Permutation Group"?

The "Number of Elements of Order 5 in S7 Permutation Group" refers to the total number of elements in the Symmetric Group S7 that have an order of 5. In other words, these are the elements that require 5 iterations to return to their original state when performing a permutation.

2. How is this number calculated?

The number of elements of order 5 in S7 can be calculated using the formula n!/k, where n is the total number of elements in the group (which is 7! for S7) and k is the order of the element (in this case, 5). So, the formula would be 7!/5 = 5040/5 = 1008.

3. Why is this number significant?

The number of elements of order 5 in S7 is significant because it represents a key property of the group - the number of elements with a specific order. This information can be used to understand the structure and properties of the group, and also has applications in various fields such as cryptography and group theory.

4. What is the relevance of S7 in this context?

S7, or the Symmetric Group of degree 7, is a group that consists of all possible permutations of 7 objects. It is one of the most studied and well-known permutation groups, and its properties are used in various mathematical and scientific fields. In this context, S7 serves as the group in which the number of elements of order 5 is being calculated.

5. Are there any other permutation groups where this calculation can be applied?

Yes, the formula for calculating the number of elements of a specific order in a permutation group can be applied to any permutation group. However, the result will vary depending on the size and structure of the group. In general, the number of elements of order k in a permutation group of size n is given by n!/k.

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