On the way to the office today, I hear two girls talking about it on the bus. I was surprised that a girl said: "Order of -1 is 2". That's why I'm here.
I do like the lecture. It is very good (far better than anything I can produce). But, I find that is somewhat suffers from the same problem all math classes do. The material covered in this example is not necessarily overly complex. However in comparison to the previous lecture, it sort of jumped the shark in terms of complexity. Perhaps an intermediary building block or going over what each statement means in very simple manner as you have done in previous lectures. Very nice work.
just needed to point out that the exponential notation used here means the number of times you would an operation (whatever operation it may be, + or x). In abstract algebra this would mean "how many times I have to apply an operation before getting the identity" this is different from the traditional use of the notation of exponents where it indicates how many times you multiply a number by.
@@walltowall5 good point. The "power operation" here is just the iterated operation n times. for example: all elements of Z under addition are infinite order except 0, because all other elements, if adding with itself gives non-zero (identity element)
I really like this series of videos and I hope you do more videos like these. The first time I took Modern/Abstract Algebra I thought I would never be able to understand the concepts. A year later with a different professor, it became super easy and fun. I don't understand why some professors purposefully make these topics difficult. I also struggled with college geometry, real & numerical analysis. It would be nice to learn and re-learn some of these topics if you decide to make more videos :)
Professor: As a teacher of algebra, I totally enjoy your videos on abstract algebra, especially the new ones on the order of an element. Please continue to make more videos on this fascinating subject. Could you make future videos on problem-solving in abstract algebra? That is, post a few typical exercises in permutation groups, cosets, and the like; then show your viewers how to solve them on the videos. As you know, practice makes perfect in math! Many thanks. Benny Lo, California
I started learning about groups earlier today (no joke) so take my comment with a grain of salt because I'm not a trained mathematician. Order is essentially the number of times - 1 you need to run an operation on a term repeatedly to generate the identity element of the group under the prescribed operator. Said a bit more mathematically: For a group G defined under a binary operator *, e is defined as an element of a set P such that for all x member of P, x*e = e*x = x. Note that * is a generic operator that could stand for anything...it is not necessarily multiplication. Some definitions (subscripts are labels): *_{1,n-1} x_t{1,n} := x_0 *_0 x_1 *_2 x_2 *_2 ... *_n x_n : n E R y := x_t{k,k} i.e. the sequence x_t{k,k} is comprised of the same term over and over again. IF there exists a positive integer k: *_{1,k-1} y = e k is defined as the order of the term y under the operator * of G. I haven't really looked into it but this seems like a more general way to formulate the idea of order of an element of a group under an unspecified operation. For example, the group G of set R^(x) under multiplication is such that x * x * x * ... * x = x^n = e = 1 i.e. x^n = 1, so give me an x and I'll give you the number 1. For x = 1, this means n = 1, in which case the multiplication operator is not used to generate the identity element, i.e. n-1 = 1-1 = 0. For x = -1, n = 2, meaning the operator is used n - 1 = 2 - 1 = 1 time, i.e. (-1) * (-1) = 1. For x = sqrt(-1), sqrt(-1) * sqrt(-1) * sqrt(-1) * sqrt(-1) = 1 i.e. the order of sqrt(-1) is 4, or number of operators used plus 1. For a Group W of set R under addition, it follows that: x * x * x * ... * x := x + x + ... + x = n * x, and we already know that the identity element of this group is 0. So n*x = 0 and if n is the order of a term x of G, it follows by similar logic that n-1 operators are needed to go from x to 0. So really, the order value of a term is just a way of sensing the number of operations it will take to run a sequential algorithm. Is this the right way to think about things? Thanks, and awesome videos! They (your videos on abstract algebra) make me wish I studied math formally lol, but knowledge is knowledge, amirite? :) I'm hoping to use this and set theory to understand the construction of the real numbers and later the hyperreal numbers. I guess you could say I'm learning to count haha. Cheers! -Float Circuit.
The order of a group is just how many elements are in the group. For example, the order of the group G of integers mod 6 is 6. You write this as |G|=6. For another example, the group of all integers under addition has infinite order. The order of an element x is the smallest positive integer n so that n*x = 0, if you are using additive notation. If you are using multiplicative notation, then it's the smallest positive integer n where x^n=1. For example in the integers mod 6, the order of 2 is 3 since 2+2+2=3*2=0 (mod 6). The order of 5 is 6, since 5+5+5+5+5+5 = 6*5 = 0 (mod 6). This happens a lot in math. The same word will be used over and over, but mean different things in different contexts.
@@Socratica You guys should consider maybe a second video on this topic highlighting the relation between the two types of order in groups of finite order. You could combine the idea of cyclic groups and Lagrange's theorem covered earlier in the playlist to observe that the set of the "powers" of an element forms a cyclic subgroup with group order equal to the order of the element. And then by Lagrange's, the order of that subgroup must divide the order of the larger group.
In Abstract Algebra we generalize numbers and operators However to express "order" we need an operator "power" which is derived from multiplication operator which is again a specialization not generalization.
Ashwin Ramaswamy you are exactly right. This video seems like misleading because in this context, 2^3 doesn't mean that 2^3=8 , it means that we are applying the given operation 3 times on element 2 to get identity element.
thanks ........these videos help me to understand the basic concept in group theory......and the method of teaching is awesome....thanks once again and love
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Thank you! And more videos are on the way. We're resuming work on this series and will cover topics like homomorphisms, isomorphisms, permutation groups and more.
I'm a bit confused about the section on roots of unity. If z^n=1, for a positive integer n, aren't there n roots? What does the infinite in "infinite z where z^n=1" refer to?
For a fixed n, there are only n complex numbers z so that z^n = 1. But, there are infinitely many z so that z^n = 1 for _some_ n. In this case, we are allowing n to vary. So you're taking all square roots and cube roots and fourth roots, etc. Both of these descriptions form a multiplicative group.
####Excellent. Will u ans. This que.? Q)Order of every element of " Group G" is equal to the order of "G"? As we know that:order of G is nothing but total no. Of elements present in G. Pls rply ma'm wtng here.
I have a question about complex numbers. I don't understand why, as stated in the video, there are infinite z where z^n = 1 with n positive integer! Please, can you help me? Thanks!
there are infinite amount of complex number that when raised to a specific power will give you 1 just to name three there is z = 1, n = 1 which is 1^1 = 1. there is z = -1, n = 2 where -1^2 = 1 then there is z = i, n =4 where i^4 = 1 and im gonna assume there exist complex numbers which raised to the power of some number n, equals 1.
i to the zero power is definitely 1. 0 to the zero is not defined (in a field). Eric is correct. For any nth root ('z') of 1, there exists an integer 'n' such that z^n=1. Think of it like this: -1^2 = 1. What is the square root of 1? +1 AND -1. That is to say, what, taken to the 2nd power is equal to one. Same with the fourth root of 1. In that case, there are 4 answers; 1, -1, i, and -i. 1 is order1, -1 is order 2, +/- i are order 4. But what about the cube root of 1? For Complex Numbers, the cube roots of 1 are: (-1/2 + root 3/2 * i) , (-1/2 - root 3 / 2 *i), and 1 itself. So, in fact, for every nth root of 1, there are n solutions. It is in this way that the Complex numbers can have infinite order. Here is the pretty part: All of these roots are on the unit circle, spaced evenly. What I mean is, each of the nth roots are evenly spaced around the circle so as to form a polygon with the 'top' oriented towards 1 on the "Real" line. The numbers themselves are the vertices of the polygon, because Complex Numbers are often expressed as trigonometric angles. That means, for example, if you want to find the 5th roots of 1, they are all on the unit circle at 0 times 2pi/5, 1 times 2pi/5, 2 times 2pi/5....etc. The sin and cosine values of these 5 angles, when raise to the 5th power, will equal 1. Test it, if you like. It's f'kin magic.
We try to keep this playlist in order: bit.ly/AbstractAlgebra but we will also be adding more example videos interspersed on our website, socratica.com. We have a page fore each topic (www.socratica.com/subject/abstract-algebra). Here is the signup for our email group, where we will notify people when new videos go live to our website: www.socratica.com/email-groups/abstract-algebra
Very precise and to the point! Its a great supplement material. By the way you are one of the prettiest lecturer I've seen...almost distracting... : ) just kidding. Great job!!!
Hi Socratica, thanks for your video! I have gotten a question here, do we have any shortcuts to calculate the order of the last matrix example that you presented in this video? just wondering if we do 12 times of matrix multiplications, that would be quite time consuming tho Thank you in advance:D
Very good question! There are definitely shortcuts. You learn about these in Linear Algebra. For many matrices M, you can write it as a product A*D*A^-1, where D is a diagonal matrix (only has numbers along the diagonal.) If you square this, you get: (A*D*A^-1)(A*D*-1) = A*D^2*A^-1. Similarly, cubing it gives you A*D^3*A^-1. In general, raising it to the nth power gives you A*D^n*A^-1. How is this helpful? Raising a diagonal matrix to the nth power is super fast: just raise all the numbers along the diagonal to the nth power. To make a long story short, for many matrices you can write them in different forms to speed up computation. For example, I raised the matrix at 2:38 to the millionth power, and it took my computer less than a second.
Also just as a sidenote, the specific matrix in the video represents a rotation. You can work out the angle it rotates by, because 2x2 matrices of the form [cosx -sinx; cosx sinx] rotate the plane by x radians. If you know your common values of cos and sin, you can see the matrix in the video is a rotation by (pi/6), so you'd have to apply it 12 times to get to 2pi (a full rotation).
You can also rule out finite order of a matrix if you can see its determinant isn't 1 or a root of unity, since determinants of matrices are multiplicative (det AB = det A * det B)
Since the real numbers under multiplication can be raised to the power of 0 to get 1, does this mean in abstract algebra zero is not a positive number?
Can you have a finite group with elements with infinite order? Ex: {e, a, b, c} where a, b, and c cycle, so a+a = b, b+a = c, c+a = a, etc? Edit: Add inverses, assuming they follow the same pattern a⁻¹+a⁻¹ = b⁻¹, b⁻¹+a⁻¹ = c⁻¹... Edit 2: This is an invalid group, I just tried making the Cayley table and ran into problems with x+y⁻¹. Sorry I should have checked first. But my question remains, is there some other formulation of such a group?
Nope! If you had an element g of infinite order, then it follows that g, g^2, g^3, g^4, etc. are all _different_ elements. If not, then there exists some j < k so that g^j = g^k. But then you could multiply both sides by g^(−j) to get e = g^(k−j), meaning that g has finite order (since k−j > 0 because j < k), which is a contradiction. So if your group has an element of infinite order, then the group has infinitely many distinct elements, meaning the group has infinite order.
I don't know any tricks generslly, but in hindsight, I see that the determinant is 1, so it preserves lengths, and must be a rotation. Since the elements are well-known values of.cosine and sine, I can see that the angle of rotation is pi/6. From that I get the rotation has to be applied 12 times to go all the way around the circle. I suppose groups of things other than rotations have their own peculiarities that you can use this way.
No definition given of what the nth power of an element means in group theory. I assume it means, repeated application of the group operator on the same element of the group?
hello socratica i have a problem that is...if G is a finite abelian group of odd order,prove that the product of all elements in G is equal to the identity element of G.please let me know how to solve this problem ..as soon as possible.
If a group has odd order, then every element must have odd order by Lagrange's Theorem. Now the question is this: can an element of odd order (other than the identity) be its own inverse?
what is the important of the definition of order of element ?why you say a^n=e n is the identity why not (a*a*a*a*a*a for n times therefore n is the identity ??
Why the order of the other elements of non-zero real numbers under multiplication is infinite? So if I take for example 2 to the 0th power I get 1. And that js the identity element.
Why is every other number except 1 and - 1, for real numbers excluding zero, under multiplication infinite order? I was thinking it would be zero order. 3 to the power of zero is equal to one, right? So should three have an order of zero and not infinite order? Or is it infinite because zero is not part of the group and n can only be elements of the group?
We mainly use the Adobe suite of software, especially Premiere and After Effects. In some of our newer videos we're starting to experiment a bit with DaVinci Resolve, too! We're self-taught filmmakers, so we're always learning. 💜🦉
@@maths_lover4870 Unfortunately, we have our hands full making videos for our own channel (you can see that we can only finish a video once every two weeks or so - it takes a huge amount of work for us to write, film, and edit all on our own). But we wish you good luck! You will learn so much by making your own videos, it's incredible.
Ehh... Humans and their fetish of fancy words & intricate definitions... :P Wouldn't it be simpler to say that the order of the operation is how many times you need to repeat it to get back to where you started? (or "close the cycle")
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On the way to the office today, I hear two girls talking about it on the bus. I was surprised that a girl said: "Order of -1 is 2". That's why I'm here.
I’ve no words that can describe your channel, also these videos are 9 years old and with that quality, that’s incredible.
2 years later I just happened to find these videos; they're still just as intriguing. Thanks for high quality content!
+Derrick McMillian We're so glad you've found our videos! Thanks for watching, and thank you for such a nice comment. :)
Who else is utilising their time during lockdown by watching this extraordinary playlist 😀?
✋🏽
Still using this during Covid!
I do like the lecture. It is very good (far better than anything I can produce). But, I find that is somewhat suffers from the same problem all math classes do. The material covered in this example is not necessarily overly complex. However in comparison to the previous lecture, it sort of jumped the shark in terms of complexity. Perhaps an intermediary building block or going over what each statement means in very simple manner as you have done in previous lectures. Very nice work.
just needed to point out that the exponential notation used here means the number of times you would an operation (whatever operation it may be, + or x).
In abstract algebra this would mean
"how many times I have to apply an operation before getting the identity"
this is different from the traditional use of the notation of exponents where it indicates how many times you multiply a number by.
@@walltowall5 good point. The "power operation" here is just the iterated operation n times. for example: all elements of Z under addition are infinite order except 0, because all other elements, if adding with itself gives non-zero (identity element)
I really like this series of videos and I hope you do more videos like these. The first time I took Modern/Abstract Algebra I thought I would never be able to understand the concepts. A year later with a different professor, it became super easy and fun. I don't understand why some professors purposefully make these topics difficult. I also struggled with college geometry, real & numerical analysis. It would be nice to learn and re-learn some of these topics if you decide to make more videos :)
Professor:
As a teacher of algebra, I totally enjoy your videos on abstract algebra, especially the new ones on the order of an element. Please continue to make more videos on this fascinating subject.
Could you make future videos on problem-solving in abstract algebra? That is, post a few typical exercises in permutation groups, cosets, and the like; then show your viewers how to solve them on the videos. As you know, practice makes perfect in math! Many thanks.
Benny Lo, California
Wow... I just want to say that these videos are amazing.
I started learning about groups earlier today (no joke) so take my comment with a grain of salt because I'm not a trained mathematician.
Order is essentially the number of times - 1 you need to run an operation on a term repeatedly to generate the identity element of the group under the prescribed operator.
Said a bit more mathematically:
For a group G defined under a binary operator *, e is defined as an element of a set P such that for all x member of P, x*e = e*x = x. Note that * is a generic operator that could stand for anything...it is not necessarily multiplication.
Some definitions (subscripts are labels):
*_{1,n-1} x_t{1,n} := x_0 *_0 x_1 *_2 x_2 *_2 ... *_n x_n : n E R
y := x_t{k,k} i.e. the sequence x_t{k,k} is comprised of the same term over and over again.
IF there exists a positive integer k:
*_{1,k-1} y = e
k is defined as the order of the term y under the operator * of G.
I haven't really looked into it but this seems like a more general way to formulate the idea of order of an element of a group under an unspecified operation.
For example, the group G of set R^(x) under multiplication is such that x * x * x * ... * x = x^n = e = 1 i.e. x^n = 1, so give me an x and I'll give you the number 1. For x = 1, this means n = 1, in which case the multiplication operator is not used to generate the identity element, i.e. n-1 = 1-1 = 0. For x = -1, n = 2, meaning the operator is used n - 1 = 2 - 1 = 1 time, i.e. (-1) * (-1) = 1. For x = sqrt(-1), sqrt(-1) * sqrt(-1) * sqrt(-1) * sqrt(-1) = 1 i.e. the order of sqrt(-1) is 4, or number of operators used plus 1.
For a Group W of set R under addition, it follows that: x * x * x * ... * x := x + x + ... + x = n * x, and we already know that the identity element of this group is 0. So n*x = 0 and if n is the order of a term x of G, it follows by similar logic that n-1 operators are needed to go from x to 0.
So really, the order value of a term is just a way of sensing the number of operations it will take to run a sequential algorithm.
Is this the right way to think about things?
Thanks, and awesome videos!
They (your videos on abstract algebra) make me wish I studied math formally lol, but knowledge is knowledge, amirite? :)
I'm hoping to use this and set theory to understand the construction of the real numbers and later the hyperreal numbers. I guess you could say I'm learning to count haha.
Cheers!
-Float Circuit.
please upload more videos on fields and ideals...and also on linear algebra.
these videos are awesome.
The order of x element of G is ||
what is difference between order of element of group and the order of whole group?
You are doing great job. Thanks.
The order of a group is just how many elements are in the group. For example, the order of the group G of integers mod 6 is 6. You write this as |G|=6. For another example, the group of all integers under addition has infinite order.
The order of an element x is the smallest positive integer n so that n*x = 0, if you are using additive notation. If you are using multiplicative notation, then it's the smallest positive integer n where x^n=1. For example in the integers mod 6, the order of 2 is 3 since 2+2+2=3*2=0 (mod 6). The order of 5 is 6, since 5+5+5+5+5+5 = 6*5 = 0 (mod 6).
This happens a lot in math. The same word will be used over and over, but mean different things in different contexts.
THANKS!
I did better in my test. Thank you.!
@@Socratica You guys should consider maybe a second video on this topic highlighting the relation between the two types of order in groups of finite order. You could combine the idea of cyclic groups and Lagrange's theorem covered earlier in the playlist to observe that the set of the "powers" of an element forms a cyclic subgroup with group order equal to the order of the element. And then by Lagrange's, the order of that subgroup must divide the order of the larger group.
Can you please come up with a video on Cyclic Groups.. Your videos are awesome.
In Abstract Algebra we generalize numbers and operators However to express "order" we need an operator "power" which is derived from multiplication operator which is again a specialization not generalization.
a^m represents m repetitions of the operation on a. They should have made it clear in the video tho.
Ashwin Ramaswamy you are exactly right. This video seems like misleading because in this context, 2^3 doesn't mean that 2^3=8 , it means that we are applying the given operation 3 times on element 2 to get identity element.
thanks
0:12 Reminder for future viewers: 0 isn't a positive integer.
That was the greatest video I have ever watched
thanks ........these videos help me to understand the basic concept in group theory......and the method of teaching is awesome....thanks once again and love
You really deserve millions of subscribers.....👍👍
This is best, until the better is done. Thank you very much
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Excellent explanation. Love it.
You are my favorite teacher
Any chance there will be additional videos any time soon? Say, quotient groups. These videos are extremely helpful.
Thank you! And more videos are on the way. We're resuming work on this series and will cover topics like homomorphisms, isomorphisms, permutation groups and more.
I'm a bit confused about the section on roots of unity. If z^n=1, for a positive integer n, aren't there n roots? What does the infinite in "infinite z where z^n=1" refer to?
For a fixed n, there are only n complex numbers z so that z^n = 1. But, there are infinitely many z so that z^n = 1 for _some_ n. In this case, we are allowing n to vary. So you're taking all square roots and cube roots and fourth roots, etc.
Both of these descriptions form a multiplicative group.
####Excellent.
Will u ans. This que.?
Q)Order of every element of " Group G" is equal to the order of "G"?
As we know that:order of G is nothing but total no. Of elements present in G.
Pls rply ma'm wtng here.
What is the purpose of knowing the order?
fell down about you and your lecture too tq so much
Why arent you releasing videos now? There should be more videos on Mathematics from magnificient tutors like you!
in a group (G,o) the elements a and b commute and o(a) and o(b) are prime to each other show that o(aob)=o(a).o(b)
I have a question about complex numbers. I don't understand why, as stated in the video, there are infinite z where z^n = 1 with n positive integer! Please, can you help me? Thanks!
there are infinite amount of complex number that when raised to a specific power will give you 1 just to name three there is z = 1, n = 1 which is 1^1 = 1. there is z = -1, n = 2 where -1^2 = 1 then there is z = i, n =4 where i^4 = 1 and im gonna assume there exist complex numbers which raised to the power of some number n, equals 1.
No - any number raised to the power of zero will equal 1.... X^0 = 1, for all numbers, except the first ones listed: 0 and i
i to the zero power is definitely 1. 0 to the zero is not defined (in a field). Eric is correct. For any nth root ('z') of 1, there exists an integer 'n' such that z^n=1.
Think of it like this: -1^2 = 1. What is the square root of 1? +1 AND -1.
That is to say, what, taken to the 2nd power is equal to one.
Same with the fourth root of 1. In that case, there are 4 answers; 1, -1, i, and -i. 1 is order1, -1 is order 2, +/- i are order 4.
But what about the cube root of 1?
For Complex Numbers, the cube roots of 1 are: (-1/2 + root 3/2 * i) , (-1/2 - root 3 / 2 *i), and 1 itself.
So, in fact, for every nth root of 1, there are n solutions.
It is in this way that the Complex numbers can have infinite order.
Here is the pretty part:
All of these roots are on the unit circle, spaced evenly. What I mean is, each of the nth roots are evenly spaced around the circle so as to form a polygon with the 'top' oriented towards 1 on the "Real" line. The numbers themselves are the vertices of the polygon, because Complex Numbers are often expressed as trigonometric angles. That means, for example, if you want to find the 5th roots of 1, they are all on the unit circle at 0 times 2pi/5, 1 times 2pi/5, 2 times 2pi/5....etc. The sin and cosine values of these 5 angles, when raise to the 5th power, will equal 1.
Test it, if you like.
It's f'kin magic.
you are the best teacher .
Good for understanding the order of the element.
Do you have a recommended order to watch these videos?
We try to keep this playlist in order: bit.ly/AbstractAlgebra but we will also be adding more example videos interspersed on our website, socratica.com. We have a page fore each topic (www.socratica.com/subject/abstract-algebra). Here is the signup for our email group, where we will notify people when new videos go live to our website:
www.socratica.com/email-groups/abstract-algebra
Very precise and to the point! Its a great supplement material. By the way you are one of the prettiest lecturer I've seen...almost distracting... : ) just kidding. Great job!!!
Now I understand what the order of an element is, but what is the motivation for defining the order of elements?
Nice explanation!!
Hi Socratica, thanks for your video! I have gotten a question here, do we have any shortcuts to calculate the order of the last matrix example that you presented in this video? just wondering if we do 12 times of matrix multiplications, that would be quite time consuming tho
Thank you in advance:D
Very good question! There are definitely shortcuts. You learn about these in Linear Algebra. For many matrices M, you can write it as a product A*D*A^-1, where D is a diagonal matrix (only has numbers along the diagonal.) If you square this, you get: (A*D*A^-1)(A*D*-1) = A*D^2*A^-1. Similarly, cubing it gives you A*D^3*A^-1. In general, raising it to the nth power gives you A*D^n*A^-1. How is this helpful? Raising a diagonal matrix to the nth power is super fast: just raise all the numbers along the diagonal to the nth power.
To make a long story short, for many matrices you can write them in different forms to speed up computation. For example, I raised the matrix at 2:38 to the millionth power, and it took my computer less than a second.
+Socratica that's super clear! thanks for your explanation! Appreciate:)
Also just as a sidenote, the specific matrix in the video represents a rotation. You can work out the angle it rotates by, because 2x2 matrices of the form [cosx -sinx; cosx sinx] rotate the plane by x radians. If you know your common values of cos and sin, you can see the matrix in the video is a rotation by (pi/6), so you'd have to apply it 12 times to get to 2pi (a full rotation).
You can also rule out finite order of a matrix if you can see its determinant isn't 1 or a root of unity, since determinants of matrices are multiplicative (det AB = det A * det B)
love your videos so much.
Thank you so much.The speech is so clear.
Since the real numbers under multiplication can be raised to the power of 0 to get 1, does this mean in abstract algebra zero is not a positive number?
hello soratica i love soo much your videos and i want more videos you are my best professer
Thank you for your kind message! Be well, Socratica Friend! 💜🦉
Can you have a finite group with elements with infinite order?
Ex: {e, a, b, c} where a, b, and c cycle, so a+a = b, b+a = c, c+a = a, etc?
Edit: Add inverses, assuming they follow the same pattern a⁻¹+a⁻¹ = b⁻¹, b⁻¹+a⁻¹ = c⁻¹...
Edit 2: This is an invalid group, I just tried making the Cayley table and ran into problems with x+y⁻¹. Sorry I should have checked first. But my question remains, is there some other formulation of such a group?
Nope! If you had an element g of infinite order, then it follows that g, g^2, g^3, g^4, etc. are all _different_ elements.
If not, then there exists some j < k so that g^j = g^k. But then you could multiply both sides by g^(−j) to get e = g^(k−j), meaning that g has finite order (since k−j > 0 because j < k), which is a contradiction.
So if your group has an element of infinite order, then the group has infinitely many distinct elements, meaning the group has infinite order.
this is better than my university :/
In the last example of matrix multiplication group, it is very difficult to get M12. Is there any tricks to solve this?
Thanks a lot for the video.
I don't know any tricks generslly, but in hindsight, I see that the determinant is 1, so it preserves lengths, and must be a rotation. Since the elements are well-known values of.cosine and sine, I can see that the angle of rotation is pi/6. From that I get the rotation has to be applied 12 times to go all the way around the circle.
I suppose groups of things other than rotations have their own peculiarities that you can use this way.
@@Rsharlan3 okay now I understand, what's going on. Its quite helpful in these kind of examples. Thanks for your help.
No definition given of what the nth power of an element means in group theory. I assume it means, repeated application of the group operator on the same element of the group?
Amazing video
Quality 👍🏻👍🏻👍🏻👍🏻
Miss you for complete course of linear algebra madam
why isn't M^6 also considered as identity metrix since it could be described as (-1)*I
great lecture😊
how is the order of the element is equal to the order of the cyclic subgroup generated by the same element
Hello teacher! Can I get pdf of this lecture?
Does order have to be a positive number? I guess the fact that a^0=1 does not make |a|=0?
Correct, order has to be positive.
hello socratica i have a problem that is...if G is a finite abelian group of odd order,prove that the product of all elements in G is equal to the identity element of G.please let me know how to solve this problem ..as soon as possible.
If a group has odd order, then every element must have odd order by Lagrange's Theorem.
Now the question is this: can an element of odd order (other than the identity) be its own inverse?
So, what if we have integers under addition? Would the only finite element be 0?
Looks like a yes to me.
Thank you so much.
Hello, can someone tell me, why is the last example a rotation matrix?
because it represents matrix:
(cos(30) -sin(30), sin(30) cos(30)) ?
what is the important of the definition of order of element ?why you say a^n=e n is the identity why not (a*a*a*a*a*a for n times therefore n is the identity ??
Why the order of the other elements of non-zero real numbers under multiplication is infinite? So if I take for example 2 to the 0th power I get 1. And that js the identity element.
Never mind, I get it. n has to be positive integer.
Determinant and the order of M in GL_n(S) would have the same sign isn't it?
Why it's called infinite order?
It means, X^(infinity)=1
thanks mam, why finite group's all element are not distinct?
This was very helpful, thank you!
I am particularly particular so thanks!
thanks u I wishes you good health
What a nice voice.
I d call it undefined and not infinite order for elements which are not 1 or -1
thanks.what is Klein's 4 grp.
thank you madam..........
so much helpful thanks!
Why is every other number except 1 and - 1, for real numbers excluding zero, under multiplication infinite order? I was thinking it would be zero order.
3 to the power of zero is equal to one, right? So should three have an order of zero and not infinite order? Or is it infinite because zero is not part of the group and n can only be elements of the group?
Nevermind. I think I figured it out haha. N is defined as "the smallest positive integer" and zero is neither positive or negative.
Thanks mam.
How you edit your vedios
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@@Socratica wow so kind to reply on a random comment on a video which is 8 years old and twice🙏😇
is the order of i not 2?
No 1 sqaure is -1 and -1 square is 1 so
Amazing
Excellent^3
Watching in 2022
هو ده التعليم ولا بلاش
Wow, this video had about a Brazillion examples!
Love from Pakistan
I like
I'm sorry. I couldn't really follow.
Ehh... Humans and their fetish of fancy words & intricate definitions... :P
Wouldn't it be simpler to say that the order of the operation is how many times you need to repeat it to get back to where you started? (or "close the cycle")