Question

Let (alpha) = (1 4 6 3 5 2) and element of S6.

a) List the elements (each as a product of disjoint cycles) of the cyclic group

Answer 1

@KingGeorge

Answer 2

I have a theorem here saying "Every permutation of a finite set can be written as a cycle or as a product of disjoint cycles" what does this mean

Answer 3

That just means that you can write the permutation in the form (i j ... k)...(x y ... z) where each of these is some disjoint cycle.

Answer 4

I.e., \[i\mapsto j\mapsto...\mapsto k \mapsto i\]\[x\mapsto y\mapsto...\mapsto z \mapsto x\]

Answer 5

As for your question at the top, you need to find all permutations generated by\[<(146352)>\]and then write each of those elements (there are 6) in cycle notation.

Answer 6

omg my internet just froze and i just fixed it sorry about that @KingGeorge

Answer 7

wow okay so could you do an example on the side because i am still confused about making the disjoint set

Answer 8

Let's just work on this problem one step at a time. We're given the element \((146352)\in S_6\). We need to find the elements that it generates. So we start multiplying it by itself.

Answer 9

so basically (146352)*(146352) ?

Answer 10

Precisely. Could you now find what that is?

Answer 11

hmm ok so wth this is confusing could you do the first one

Answer 12

but what iam getting is (1,4,6)

Answer 13

then i try doing 2 and it goes back to 4

Answer 14

so i think i have a wrong start

Answer 15

Alright. Start on the RIGHT when you do it this way. Follow the path that each number takes individually. First in the permutation on the right, and then on the left \[1\mapsto4\mapsto6\]\[2\mapsto1\mapsto4\]\[3\mapsto5\mapsto2\]\[4\mapsto6\mapsto3\]\[5\mapsto2\mapsto1\]\[6\mapsto3\mapsto5\]Now we ignore the middle column we get \[1\mapsto6\]\[2\mapsto4\]\[3\mapsto2\]\[4\mapsto3\]\[5\mapsto1\]\[6\mapsto5\]Now we have to write this in cycle notation. This is the easy part. You should just get \[(165)(243)\]Do you understand where I got that from?

Answer 16

ooooo i see i see i was including the middle column for mine

Answer 17

yeah that makes so much sense when you write it like that

Answer 18

so is this what we do whenever we are given a permutation like that we multiply it by itself?

Answer 19

Whenever you multiply permutations, this is the most surefire way to get the solution. There are other methods, but I usually find this the easiest.

Answer 20

yeah it is really easy now do we multiply by it self again or is this enough

Answer 21

We need to keep multiplying by \((146352)\) until we reach the identity. Can you tell me what \[(146352)\circ(146352)\circ(146352)=(165)(243)\circ(146352)\]is?

Answer 22

I've got to go get dinner now and leave for a while :/

I'll be back in an hour or so if you want to check this stuff with me. Just post all the elements generated by \((146352)\) as you can.

Answer 23

alright awesome thanks a lot i will be on here :) i will just post another question later then go home but i will continue finding this right now but for your question i got (13)(26)(45)

Answer 24

Looks perfect. Just continue like that.

Answer 25

and if i may ask quickly what is the identity lol

Answer 26

(1)(2)(3)(4)(5)(6)

Answer 27

ooo sick alright thanks alot i shall talk to you soon again

Answer 28

for (13)(26)(45)*(146352) i got (156)(234)

Answer 29

ok sick thank god i didnt give up i got my final answer as when alpha^6 = (1)(2)(3)(4)(5)(6) :D

Answer 30

alright so @kingGeorge when you are back this is what i got so is the answer to my question as these are the elements of alpha?

part b to that question says that

B) List the subgroups and generators of <alpha>

Answer 31

what does that mean :s

Answer 32

Right, just to list them all out here, we have \[(146352)^1=(146352)\]\[(146352)^2=(165)(243)\]\[(146352)^3=(13)(26)(45)\]\[(146352)^4=(156)(234)\]\[(146352)^5=(125364)\]\[(146352)^6=(1)(2)(3)(4)(5)(6)\]This is everything that's generated by \((146352)\) since \((146352)^7=(146352)\). Now we need to find the subgroups and generators of this group.

Here's a quick trick to find the generators of a cyclic group. Once you have all the elements written in the form of \(g^k\) (here \(g=(146352)\)), find those with an exponent relatively prime to \(n\) (here \(n=6\)). That means that our generators will be\[\alpha=(146352)\]\[\alpha^5=(125364)\]

Answer 33

Now the subgroups. This is also easy for cyclic groups. Instead of taking those exponents relatively prime to 6, take those that are not relatively prime to 6, and find the subgroups those generate. You may be thinking, "but how do we find what they generate? We'll have to do this all over again!" I'm glad to tell you we don't have to do this, we just have to add the exponents until we get to something divisible by \(n=6\). That means we will have the following subgroups.

\[\alpha^6=(1)(2)(3)(4)(5)(6)=e \qquad \text{(Let's just call this e for now)}\]\[<\alpha^2>=<\alpha^4>=\{e,\;(165)(243),\; (156)(234)\}\]\[<\alpha^3>=\{e,\;(13)(26)(45)\}\]

Answer 34

hey @KingGeorge i am back i am going to look at this noww

Answer 35

okay so you know how you found those generators how did you actually find that did you do the generators of U(6) ?

Answer 36

like why are the generators alpha and alpha^5?

Answer 37

because U(6)={1,5}

Answer 38

\(\alpha\) and \(\alpha^5\) are the generators merely because 1 and 5 are relatively prime to 6, and this is a cyclic group.

Answer 39

is that a theorem?

Answer 40

Yes, I'm not sure if it has a specific name, but this is definitely true, and not terribly difficult to prove.

Answer 41

alright so basically if Un is a cyclic group then so are the subgroups relative prime to n basically?

Answer 42

Sort of. If Un is a cyclic group, then ALL subgroups are cyclic.

Answer 43

If you have a generator for Un, then the powers of that generator that are coprime to n, are also generators.

Answer 44

yeah true your right

Answer 45

i have that in my book so now for the subgroups

Answer 46

ok you know how you said "we just have to add the exponents until we get to something divisible by n=6." what do you mean by adding exponenets

Answer 47

like i dont get how you put alpha^2 = alpha^4

Answer 48

Let's start with \(\alpha^2\). The subgroup generated by this will consist of\[\alpha^2,\;\;\alpha^{2+2}=\alpha^4,\;\;\alpha^{4+2}=\alpha^6\]Since we've reached a power of 6, we stop.

Also, if we look at \(\alpha^4\) instead, we get that the subgroup will have \[\alpha^4,\; \;\alpha^{4+4}=\alpha^8=\alpha^2,\;\; \alpha^{2+4}=\alpha^6\]Once again, we stop since this is divisible by 6. From this, we can tell that the subgroup generated by \(\alpha^2\) is the same as the subgroup generated by \(\alpha^4\). Hence, we can say that \[<\alpha^2>=<\alpha^4>\]

Answer 49

and for alpha^3 = alpha^(3+3) = alpha^6 and we stopped since we reached 6

Answer 50

Exactly.

Answer 51

and if i may ask why did we choose the numbers that WERENT relative prime to 6?

Answer 52

Because the ones that do generate the entire group. If we want to generate a proper subgroup, we need to choose a generator that doesn't generate the entire group.

Answer 53

However, you should add the whole group as another subgroup, giving us \[\{e\}\]\[<\alpha>\]\[<\alpha^2>\]\[<\alpha^3>\]as a total list of subgroups.

Answer 54

what happened to alpha^4?

Answer 55

so arent my subgroups just

<α2>=<α4>={e,(165)(243),(156)(234)}
<α3>={e,(13)(26)(45)}

Answer 56

I skipped \(\alpha^4\) and \(\alpha^5\) since \[<\alpha>=<\alpha^5>\]\[<\alpha^2>=<\alpha^4>\]

Those are your proper subgroups. If you want a complete list, you have to include the identity subgroup, and the whole subgroup as well.

Answer 57

ahh i see so my subgroups are <e>, <alpha>, <alpha^2>,<alpha^3> and my generators are simply <alpha> and <alpha^5>

Answer 58

Correct.

Answer 59

alright awesome that made quite a sense i am going to look at the other question but before that this problem is troubling me

Answer 60

so this is completely off topic now say we have U(11) correct and it says find the generators
i know it is 2 6 7 and 8 but i dont feel like doing the long process asdoing

2^1 mod 11, 2^2 mod 11, ....,2^10 mod 11= 1 and this goes for 6 7 and 8

Answer 61

is there any simpler way, i already did the question by doing that method but just thinking for tests or exams

Answer 62

like what i was thinking cant i just do 2^10 mod 11 and 6^10 mod 11 and if it gives me 1 then it is a generator?

Answer 63

You don't have to go all the way to \(2^{10} \pmod{11}\). You merely need to calculate up to \(2^{5}\pmod{11}\). If \(2^k\cancel{\equiv} 1\pmod{11}\) for all \(k\leq 5\), then 2 is a generator.

Answer 64

yeah there was a theorem like that but i didnt know if the teacher would be like no

Answer 65

Unfortunately, you can't just do \(2^{10}\pmod{11}\) since \(2^{10}\equiv3^{10}\equiv4^{10}\equiv...\equiv10^{10}\equiv 1\pmod{11}\)

Answer 66

Now that you know that 2 is a generator however, you can find the other generators quickly. Since 1, 3, 7, 9 are relatively prime to 10, you know that \[2^1\equiv2\pmod{11}\]\[2^3\equiv8\pmod{11}\]\[2^7\equiv9\pmod{11}\]\[2^9\equiv6\pmod{11}\]are all of the other generators.

Answer 67

yeah true lol so if i use that theorem for every element in U(11) and if it isnt equal to 1 it is a generator

Answer 68

WOH how did you do that lol that fast

Answer 69

o my god your right there was a theorem like that by looking at the primes relative prime to phi(11) and those would be it

Answer 70

your smart thanks a lot man

Answer 71

You're welcome. If you want some more information on this stuff, you're looking for numbers that are called "primitive roots." There's plenty more information about this topic on wikipedia and wolfram http://en.wikipedia.org/wiki/Primitive_root_modulo_n

Answer 72

yeah generators and primitve roots are the same thing right?

Answer 73

If you're talking about using a subset of the integers with multiplication, they are the same thing. Primitive roots are a specific kind of generator.

Answer 74

true truee alright i am going to look at the other question noww and after that i only have one more question left

Answer 75

would you be leaving anytime soon?

Answer 76

I'll be on for a little bit more. Not too much longer however.

Answer 77

alright i will just write the qeustion and i will interrupt you if i get stuck on the other solution, but anywho for the other one you know how i was getting 2 or 3 columns whatever the last number is that is what we want correct??

Answer 78

Correct.

Answer 79

and my other question is

A)Find three elements in S7 of order 6 such that each has a different cycle type?

B)How many elements in S9 and order 10?

Answer 80

i dont know how to do A but for B i got t as there are no 10-cycles in S9, they'd have to be the products of disjoint 5- and 2- cycles. There are 9p5 5-cycles times 4p2 or 12 2-cycles (transpositions), so their product would be it probably wrong but i dont know tried lol

Answer 81

A) Just find three different cycle types and then fill it in with the easiest permutation.
-Cycle Type 1: Every cycle has order 1. Then your element would be \((1)(2)(3)(4)(5)(6)(7)\).
-Cycle Type 2: 5 cycles of order 1, 1 cycle of order 2. An element would be \((1)(2)(3)(4)(5)(67)\).
-Cycle Type 3: 4 cycles of order 1, 1 cycle of order 3. An element would be \((1)(2)(3)(4)(567)\).

Answer 82

For B, note that it's asking for the order of the element. We don't want 10-cycles. However, we do want something with a cycle type of 2,5,1,1 or 2,5,2. So we need to count the permutations with those cycle types.

Answer 83

The number of permutations with cycle type 2, 5, 1, 1 is given by the formula \[\binom{9}{2}\cdot\binom{7}{5}\cdot\frac{5!}{5}\]And the number of permutations of cycle type 2, 5, 2 is the same thing. So you should have exactly \[2\cdot\binom{9}{2}\cdot\binom{7}{5}\cdot4!\]I'll explain how I got that in the next post.

Answer 84

First, we need to choose two elements from {1, 2, 3, 4, 5, 6, 7, 8, 9} to be part of the 2-cycle. We get \(\binom{9}{2}\).

Next, we need to choose five elements to be in a 5-cycle. We can have \(\binom{7}{5}\) choices of those 5 elements, but we have to find the different orderings. That means permuting those 5 elements, so we multiply by \(5!\). However, (12345)=(23451), so we have to divide by 5.

Finally, we need to choose two more elements to remain fixed. So we multiply by \(\binom{2}{2}=1\). Thus, we get \[\binom{9}{2}\cdot\binom{7}{5}\cdot\frac{5!}{5}\]

Answer 85

For the cycle type 2, 5, 2, it's the same until the last part. It's just we have \(\binom{2}{2}=1\) because we want to choose two elements to swap.

Answer 86

However, I just realized I forgot one more cycle type. We also need to include the cycle type 4, 5.

Answer 87

alright so let me ask some questions because i am kind of confused on some stuff should i start with this question or the one in our messages?

Answer 88

Let's start with this one.

Answer 89

alright ok so you know for A we are told to find of order 6

Answer 90

do we just make our own because you did of order 1 order 2 and order 3?

Answer 91

or is it the same thing :s

Answer 92

I screwed that one up didn't I? Good thing you told me about that.

In that case, we'll merely use the cycle types
1) 2,3,1,1
2) 2,3,2
3) 6,1

I'll let you put the numbers in for the permutations you need.

Answer 93

but where are you getting those numbers from that was going to be question for B where you chose 2,5,1,1 and 2,5,2

Answer 94

Basically, (for A) we're looking for a set of numbers that add to 7, and have a least common multiple of 6.

For B, we're looking for a set of numbers that add to 9, and have a lcm of 10.

Answer 95

OO ok ok A makes sense now

Answer 96

but how does the permutation look like for (16)

Answer 97

is it just (123456)(7)?