I previously had a video on this, but I used a needlessly complicated definition which made the lesson a few minutes longer than it needed to be. So I redid it.
Thanks for the videos; they are immensely helpful!!! Question about the example presented at 4:40: The way I look at it is: Cycle 1: Start with 1: 1 --> 6 (1) 6 --> 4 (2) 4 -> 2 (3) 2 -> 1 (4) I started with 1 and ended with 1 so to get back to 1, I had to make 4 “jumps” so 4 is a possible answer for our order. Cycle 2: Skip all the numbers covered in Cycle 1 so I start with 3. 3 --> 3 (0) Since 3 maps to itself, I do not consider a “jump”. 0 would be a possible answer but since it is non-positive, I can eliminate it as a possibility. Cycle 3: 5 --> 5 (0) Again, not a “jump” because 5 maps to itself. Again, 0 can be eliminated as a possibility since it is non-positive. So that is why the order is 4. Is this way of thinking correct or is there something important I’m missing? Thank you!!
I was wondering why not 0 for the order, and the reason it is not 0 is because every element to the 0th power is the identity element, so it has to be a positive integer to be non trivial
If e is the identity element of G, and e is also a power of some a ∈ G, then can we say ord(e) = 0, or is it ord(e) = 1 because the order of an element is strictly for non-zero positive integers? Edit: Nevermind I watched the next video
Hi W.O.M Would love some videos on ramsey theory! Could be a great fit in your graph theory playlist, or just within a general combinatorics theme. There is definitely room for a more intuitive explanation on RUclips. Love the Abstract algebra vids.
The abuse of notation really doesn't help in this video. I was wondering why you suddenly switched from saying the order is "n" where "a * n = e", to saying "a^n = e". Those are two completely different operations, multiplication vs. exponentiation! But, then I realized that Group Theory applies generally to any set with an associated binary operation operation. So, in the beginning, when you said a * n = e, you were talking about Groups with the Addition operation +. So, using the proper notation for Groups, for the first example, where you wrote 3 ∈ ℤ6, you really meant it belonged to the additive group 3 ∈ (ℤ6, +). In this case, we are asking how many copies of 3 do we need to get to the identity, but because the operation is addition, we are asking how many times do we have to ADD 3 together to get the identity, which is the same as asking what do we have to multiply by 3 to get the identity. But, in the case you were saying the order of the element "a" was "n" where "a^n = e", you were talking about Groups with the multiplication operation, written as (G, *) . So, you are still asking the question "How many copies of 3 do I need to combine to get the identity", but since the operation in this case is multiplication, we are asking how many times do we have to MULTIPLY 3 together to = e. The number of times we multiply a number is the same as taking the exponent to that power. Please let me know if I am right about this, I am still learning this subject myself.
Isn't a group technically a set of ordered pairs? So it is wrong to say that 3 is an element of (Z6,+). 3 is an element of Z6 which is a group under addition. Also I believe * is reserved for arbitrary operations while multiplication preserves its dot symbol. Also when he was discussing examples in order he said he wrote them in multiplicative notation; not necessarily referring to the operations multiplication or addition specifically. This I believe just means a*b*c. Regarding powers (a^n), I believe that they are raised to n under the operation of the group. I.e if the operation was o then a^2=a o a. But I'm only in 9th grade so I might as well be even more wrong than you are.
am i the only one who couldn't understand this 😭😭? I'm feeling very dumb, i watched it so many times and still can't get it well idk why🥹!! ( I'm a math major)
I previously had a video on this, but I used a needlessly complicated definition which made the lesson a few minutes longer than it needed to be. So I redid it.
Thanks for the videos; they are immensely helpful!!!
Question about the example presented at 4:40:
The way I look at it is:
Cycle 1:
Start with 1:
1 --> 6 (1)
6 --> 4 (2)
4 -> 2 (3)
2 -> 1 (4)
I started with 1 and ended with 1 so to get back to 1, I had to make 4 “jumps” so 4 is a possible answer for our order.
Cycle 2:
Skip all the numbers covered in Cycle 1 so I start with 3.
3 --> 3 (0)
Since 3 maps to itself, I do not consider a “jump”.
0 would be a possible answer but since it is non-positive, I can eliminate it as a possibility.
Cycle 3:
5 --> 5 (0)
Again, not a “jump” because 5 maps to itself.
Again, 0 can be eliminated as a possibility since it is non-positive.
So that is why the order is 4.
Is this way of thinking correct or is there something important I’m missing?
Thank you!!
Thanks sir , I try to improve math skills by your video
6:47 will the order be 3? because you are starting with 1->6 but i think we included it in while counting
Good explanation! Keep going, You do good job.
Thanks, will do!
thanksssss😭💘💘💘💘
You can never go wrong with Wrath of Math!
Such is the order of things!
The frog is the icing on the cake.
It always is!
Thank you.
Glad to help!
So helpful thanks
Thanks for watching!
Thank u so much 🖤
You're welcome!
I was wondering why not 0 for the order, and the reason it is not 0 is because every element to the 0th power is the identity element, so it has to be a positive integer to be non trivial
If e is the identity element of G, and e is also a power of some a ∈ G, then can we say ord(e) = 0, or is it ord(e) = 1 because the order of an element is strictly for non-zero positive integers?
Edit: Nevermind I watched the next video
very interesting
I think so too!
i love you
Hi W.O.M
Would love some videos on ramsey theory!
Could be a great fit in your graph theory playlist, or just within a general combinatorics theme.
There is definitely room for a more intuitive explanation on RUclips.
Love the Abstract algebra vids.
The abuse of notation really doesn't help in this video. I was wondering why you suddenly switched from saying the order is "n" where "a * n = e", to saying "a^n = e". Those are two completely different operations, multiplication vs. exponentiation! But, then I realized that Group Theory applies generally to any set with an associated binary operation operation. So, in the beginning, when you said a * n = e, you were talking about Groups with the Addition operation +. So, using the proper notation for Groups, for the first example, where you wrote 3 ∈ ℤ6, you really meant it belonged to the additive group 3 ∈ (ℤ6, +). In this case, we are asking how many copies of 3 do we need to get to the identity, but because the operation is addition, we are asking how many times do we have to ADD 3 together to get the identity, which is the same as asking what do we have to multiply by 3 to get the identity. But, in the case you were saying the order of the element "a" was "n" where "a^n = e", you were talking about Groups with the multiplication operation, written as (G, *) . So, you are still asking the question "How many copies of 3 do I need to combine to get the identity", but since the operation in this case is multiplication, we are asking how many times do we have to MULTIPLY 3 together to = e. The number of times we multiply a number is the same as taking the exponent to that power. Please let me know if I am right about this, I am still learning this subject myself.
Isn't a group technically a set of ordered pairs? So it is wrong to say that 3 is an element of (Z6,+). 3 is an element of Z6 which is a group under addition. Also I believe * is reserved for arbitrary operations while multiplication preserves its dot symbol. Also when he was discussing examples in order he said he wrote them in multiplicative notation; not necessarily referring to the operations multiplication or addition specifically. This I believe just means a*b*c. Regarding powers (a^n), I believe that they are raised to n under the operation of the group. I.e if the operation was o then a^2=a o a. But I'm only in 9th grade so I might as well be even more wrong than you are.
am i the only one who couldn't understand this 😭😭? I'm feeling very dumb, i watched it so many times and still can't get it well idk why🥹!! ( I'm a math major)
thanksssss😭💘💘💘💘
Glad to help, thanks for watching!