Question

Let T be in S(sub7). Suppose T^3= (1,2,3,4,5,6,7), a 7-cycle. Determine what permutation T represents and justify your answer.

Answer 1

Well let's think about it. We know that T^3 has order 7, so T itself must have order dividing 3*7=21. Since it obviously can't have order 1 or 3, T must have order 7. So let's write the permutation as follows.\[1\mapsto?\mapsto?\mapsto2\]\[2\mapsto?\mapsto?\mapsto3\]\[3\mapsto?\mapsto?\mapsto4\]\[4\mapsto?\mapsto?\mapsto5\]\[5\mapsto?\mapsto?\mapsto6\]\[6\mapsto?\mapsto?\mapsto7\]\[7\mapsto?\mapsto?\mapsto1\]Each time we travel across the arrow, we're applying T. Since we've determined that T has order 7, let's follow the arrow 7 times for each number, and wee what we get.\[1\mapsto6\mapsto?\mapsto2\]\[2\mapsto7\mapsto?\mapsto3\]\[3\mapsto1\mapsto?\mapsto4\]\[4\mapsto2\mapsto?\mapsto5\]\[5\mapsto3\mapsto?\mapsto6\]\[6\mapsto4\mapsto?\mapsto7\]\[7\mapsto5\mapsto?\mapsto1\]But this all we need. We've exactly described T as \[T=(1642753)\]

Answer 2

Did the process make sense?

Answer 3

I'm not sure I follow what the areas are doing. We just went over this in class today and it made NO sense at all

Answer 4

The arrows? They just represent applying the permutation \(T\). So permutation the (1234567) could be written as
\[1\mapsto2\]\[2\mapsto3\]\[3\mapsto4\]\[4\mapsto5\]\[5\mapsto6\]\[6\mapsto7\]\[7\mapsto1\]In the work I showed above, I was using the arrows as an alternate notation for \(T\). So by moving across the arrows 3 times, we apply T three times, which is the same as applying \(T^3\).

Answer 5

ok thank you!
but how did you get (1642753)? I get that we are applying T 3 times, but how did you get that answer. I'm sorry I just really don't get this kind of thing

Answer 6

You understand why T^7 is the identity correct?

Answer 7

....not really. I know that the order of S7 is 7! and that the order of T is 7!/2 which is 504.....I think that's right

Answer 8

Alright, so the order of \(S_7\) is 7!. The order of \(T\) must therefore divide \(7!\). But you know that the order of \(T^3\) is 7 (since it's a 7-cycle). Thus, \((T^3)^7=T^{21}=e\).

Still following me?

Answer 9

yes! this makes much more sense than class

Answer 10

Excellent.

Then since \(T^{21}=e\), you know that if \(T^r=e\), and \(r\) is minimal, then \(r\) must divide 21. The only divisors of 21 are 1,3,7,21, so we only need to see if \(T\) can have order 1,3,7 or 21. You still understand this?

Oh, and btw, I just thought of a simpler way to get find \(T\), but we still need this information.

Answer 11

ok go on. I'm still with you

Answer 12

Great.

If \(T\) has order 1, then \(T=e\), and \(T^3=e\neq(1234567)\). So the order of \(T\) is not 1.

If \(T\) has order 3, then \(T^3=e\), so it's the same thing if it had order 1.

Now, this statement is the only part where I'll do a little hand-waving. There are no elements in \(S_7\) of order 21. You can check this yourself if you know how to compute the order of a general element in \(S_n\). Otherwise, just take my word for it.

Thus, \(T\) must have order 7.

Still understanding what I'm doing?

Answer 13

yes

Answer 14

Now is when I'll start to do things a bit differently.

Since the order of \(T\) is 7, we know that \(T^7=e\), so \(T^6=T^{-1}\). But we're given \(T^3\). So we can easily find \(T^6=(T^3)^2\). Can you tell me what \(T^6\) is?

Answer 15

....the inverse? I don't know

Answer 16

Think of \(T^6\) as \((T^3)^2\) and not the inverse.

Answer 17

so it's what we had originally, we just permutate....twice?

Answer 18

Precisely.

Answer 19

....so....I don't follow what you are trying to tell me from that.

Answer 20

If \(T^3=(1234567)\), what is \((T^3)^2=(1234567)(1234567)\)?

Answer 21

is this mod 7?

Answer 22

I guess you could think of it like that. It shouldn't matter though.

Answer 23

well am i multiplying them or not? This is the part that he didn't cover in class. would everything just shift one? so then its (7123456)?

Answer 24

Alright. Let's review how to multiply two permutations with a small example. Suppose the permutation we have is \((13425)\), and we want to find \((13425)(13425)\).

Let's look at where 1 is sent to. We apply (13425) to 1, and we get 3. Now if we apply (13425) again, 3 gets sent to 4.

We ended with 4, so let's see what happens when we apply (13425) to 4. Well, it gets sent to 2. Applying (13425) again we see that 2 is sent to 5.

We do it again with 5. Using the same process, 5 gets sent to 1, which is then sent to 3.

With three, it gets sent to 4, and then 2.

Finally, 2 gets sent first to 5, which then is sent 1. Thus, we can conclude that

(13425)(13425)=(14532).

Answer 25

Let me know if you have questions about that.

Answer 26

so far I have (1723456)......

Answer 27

Take a step back. Do you understand how multiplication works with permutations?

Answer 28

For example, can you find (123)(234)?

Answer 29

no we've never done it before.... I'm assuming that 1 goes to 2 3 goes to 3 and 3 goes to 4

Answer 30

so maybe (143)......I really don't know.

Answer 31

No.

First look at 1. If we apply (234) to 1, nothing happens. But then we apply (123) which sends 1 to 2.

Now we look at 2. (234) sends 2 to 3. But then we apply (123) to 3, to get 1. Since we just found where 1 was sent, we know that (123)(234) has the 2-cycle (12)

Next, look at 3. This immediately gets sent to 4, but (123) doesn't change 4.

Finally, look at 4. (234) sends 4 to 2, and then (123) sends 2 to 3. So we also have the 2-cycle (34).

Therefore, (123)(234)=(12)(34).

Answer 32

ohhhhhhhhhhhh ok. but we only want one 7-cycle a

Answer 33

as our answer right?

Answer 34

In the example we're working with, we do want to end up with a 7-cycle. So let's try one more time.

What's (1234567)(1234567)?

Answer 35

ok.... 1 goes to 2, 2 goes to 3, 3 goes to 4, 4 goes to 5, 5 goes to 6, 6 goes to 7.....7 then goes to 1 and then 2.....from there I'm not sure

Answer 36

I'm not doing a great job at how to multiply permutations. You might want to check up with your teacher about how to do it, or read through these links. They give a pretty decent explanation.
http://math.stackexchange.com/questions/31763/multiplication-on-permutation-group-written-in-cyclic-notation
http://www.mymathforum.com/viewtopic.php?f=13&t=10019

But for now, lets keep working with (1234567) step by step.

Answer 37

First. Let's look at 1. After applying (1234567) twice, what does 1 get sent to?

Answer 38

...3?

Answer 39

because 1 goes to 2 which goes to 3?

Answer 40

Perfect. What happens to 3 if you apply (1234567) twice?

Answer 41

it goes to 5?

Answer 42

Right again! Where does 5 end up? What about 7?

Answer 43

5 goes to 7.....and 7 goes to 2

Answer 44

Perfect again. Continuing this, 2 goes to 4, 4 goes to 6, and 6 goes to 1.

Putting this all together, we get that (1234567)(1234567)=(1357246).

Answer 45

HOW!? That's where I'm confused. we just said (6712345) so how are you getting (1357246)

Answer 46

oh wait......wait i get it. holy cow i see it now cause 1 goes to 3 and 3 goes to 5....etc???????

Answer 47

That's precisely it. Gotta love those A-HA moments :)

Answer 48

haha so now what? we have T^6..... I need T

Answer 49

Well we know from our discussion way above that \(T^6=T^{-1}\). I have three ways to find \(T\). A difficult way, a slightly confusing way, and a magical way :P

Which way would you prefer?

Answer 50

um....magic lol. Unicorns and rainbows and everything.

Answer 51

Magical way here we go.

Suppose you have a \(k\)-cycle \((i_1\,i_2\,...\,i_k)\). Then its inverse is \((i_1\,i_k\,i_{k-1}\,...\,i_3\,i_2)\).

Answer 52

so (1357246) goes to (1642753)?

Answer 53

Yup. So that's \(T\).

Answer 54

HOLY CRAP!

Answer 55

You are amazing! thank you thank you THANK YOU!

Answer 56

That is the most amazing thing....ever. I love math when I finally understand it. Thanks KingGeorge!!!!

Answer 57

You're welcome.

And btw, the slightly confusing would have basically proved the magical way. I'll let you work out the details for yourself if you want to.

The difficult way would have been raising \(T^{-1}\) to the 6th power. Which still isn't that bad, but it's not great.