Your welcome! Two ways to think about it: we can always start by just defining the isomorphism and checking the properties - set up two 6x6 tables and check that the multiplications match. This is not the best way to check for isomorphism, but should be done with a few small examples. A better way is to note that both groups are generated by a single element: 1 and (1,1), both of these elements have order 6, and both groups have order six. Then show pi(a^k)=pi(a)^k.
You're welcome! I'm definitely feeling the love on the Algebra videos. The plan for the near future is: finish Field Theory (less than two months?), get Precalculus going for teaching, and then Real Analysis. Francis Su from Harvey Mudd has a great real analysis course up on TH-cam, so I have to bring my A game.
hi there, the order of the elements is determined on how many times you should apply the operation to get the identity. for example when we mapped 2 to (0,2) in Z2 x Z3, (0,2) + (0,2) = (0,4) = (0,1) in Z2 X Z3 then we add again. (0,1) + (0,2) = (0,3)=(0,0) in Z2 X Z3 so three times to get to the identity so its order is 3.
Hey Dr. Bob thanks for your time in making these videos, I am sure its helping many undergrad students in math like me. I really hope by this time next year you'll have a series of real analysis videos :)
Can you answer this B. Outline a proof of the following fact: The following relation on the category of groups is an equivalence relation, partioning that Category into equivalnece classes: (G,*) ~ (H,+) if and only if there exists an isomorphism f: (G,*)-->(H,+).
They are all located in the Abstract Algebra playlist, so no need to link again. You need to point to something specific. There are many ways to use these.
When you say that G = = {...,g^-2,g^-1,e,g,g^2,...} is a cyclic group don't you mean since it is under addition we should be multiplying by some n? Or are you using for example g^2 as a placeholder for any operation?
Help with these problems please, some hints Determine (with proof) the isomorphism classes for groups of each order up to 5. Suppose φ : G→H is an onto group homomorphism (also known as an epimorphism). Prove that if G is cyclic, then H is cyclic. Let G be an abelian group. Show that the map φ(g) = g-1 ∀ g ∈ G is an epimorphism. Show that if G is not abelian, then φ is not a homomorphism.
Order 5: we know Z/5 is one class (cyclic of order 5). If we have another order 5 group H, consider powers of any nonidentity element. Onto homomorphism: suppose x generates G. show that pi(x) generates H. Abelian: show p(gh)=p(g)p(h). What is the inverse map to p? Not abelian: consider S_3.
That plan sounds great, I am definitely gonna check out Prof. Francis channel on that, thanks for the reference. Dr.Bob, just my two cents here, but don't you think its better to make more math videos in areas which do not have good series of them available on youtube already, for example I am taking complex variables right now, and haven't found too much on youtube to refer too. For things like pre-calc, many people like khanacademy have great videos on that already.
+prudhvi raj Big question. For your example, it's greatly simplified since Z/21 is cyclic. Then any homomorphism is determined by sending a generator (say 1) to an element with order that divides 21. So only the trivial homomorphism (all to 0) for Z/41, but interesting if Z/3 or Z/15, for examples.
I see that you have used two different kinds of arrows to denote a homomorphism from G to H. One uses a double-headed arrow (both heads facing the same way)and the other uses a single-headed arrow. Is there a different meaning to each of these?
Hi Bob, i have a question on the example at 8:15, Z6 ~ Z2 x Z3 . I am confused on how we can determine if they are isomorphic and how did u get the order of each element in Z2 x Z3. Thanks so much!!~~
I understand, but think of it this way: The Galois theory videos I'm doing right now are a lot of fun, but they can be a brutal slog because it is very easy to make errors. If I did only middle to upper undergraduate material, I'd burn out in no time.
a question please MathDoctorBob, Decide if the following claim is true or false, and give arguments to support your answer. The claim is that any two abelian groups of order 99 are isomorphic. I think that they are isomorphic since they are the same order and are both abelian please help.
In general, order is not enough to distinguish abelian groups. In the case of order 99, z/3 x z/33 and z/99 are not isomorphic. z/3 x z/33 doesn't have any elements of order 9 or 99 (only 1, 3, 11, and 33). The general result is the fundamental theorem of finite abelian groups.
Hi doctor bob! just wanna ask in the proof of Z6 is iso to Z2 x Z3 how does the proof of homomorphism go? thanks for the wonderful videos! I'm learning algebra easily :)
yeah me neither it feels like he's just reading of a scrip or something, like a robot. I don't feel like he even understands it himself so it's really hard to understand
!["ซี - เอมี่ " ย้อนความรัก 19 ปี ที่ต้องทนกันและกัน... | แฉ 22 ต.ค. 67 [2/3] | GMM25](http://i.ytimg.com/vi/ykv9ILzatHM/mqdefault.jpg)
Your welcome! Two ways to think about it: we can always start by just defining the isomorphism and checking the properties - set up two 6x6 tables and check that the multiplications match. This is not the best way to check for isomorphism, but should be done with a few small examples. A better way is to note that both groups are generated by a single element: 1 and (1,1), both of these elements have order 6, and both groups have order six. Then show pi(a^k)=pi(a)^k.
You're welcome! I'm definitely feeling the love on the Algebra videos. The plan for the near future is: finish Field Theory (less than two months?), get Precalculus going for teaching, and then Real Analysis. Francis Su from Harvey Mudd has a great real analysis course up on TH-cam, so I have to bring my A game.
You are the best Dr. Bob.
Thanks for helping again and for your valuable time
hi there, the order of the elements is determined on how many times you should apply the operation to get the identity. for example when we mapped 2 to (0,2) in Z2 x Z3, (0,2) + (0,2) = (0,4) = (0,1) in Z2 X Z3 then we add again. (0,1) + (0,2) = (0,3)=(0,0) in Z2 X Z3 so three times to get to the identity so its order is 3.
Hey Dr. Bob thanks for your time in making these videos, I am sure its helping many undergrad students in math like me. I really hope by this time next year you'll have a series of real analysis videos :)
Can you answer this B. Outline a proof of the following fact: The following relation on the category of groups is an equivalence relation, partioning that Category into equivalnece classes:
(G,*) ~ (H,+) if and only if there exists an isomorphism f: (G,*)-->(H,+).
thank you very much Dr. Bob. Your videos are EXCELLENT
They are all located in the Abstract Algebra playlist, so no need to link again.
You need to point to something specific. There are many ways to use these.
When you say that G = = {...,g^-2,g^-1,e,g,g^2,...} is a cyclic group don't you mean since it is under addition we should be multiplying by some n? Or are you using for example g^2 as a placeholder for any operation?
Help with these problems please, some hints
Determine (with proof) the isomorphism classes for groups of each
order up to 5.
Suppose φ : G→H is an onto group homomorphism (also known as
an epimorphism). Prove that if G is cyclic, then H is cyclic.
Let G be an abelian group. Show that the map φ(g) = g-1 ∀ g ∈ G
is an epimorphism. Show that if G is not abelian, then φ is not a
homomorphism.
Order 5: we know Z/5 is one class (cyclic of order 5). If we have another order 5 group H, consider powers of any nonidentity element.
Onto homomorphism: suppose x generates G. show that pi(x) generates H.
Abelian: show p(gh)=p(g)p(h). What is the inverse map to p? Not abelian: consider S_3.
That plan sounds great, I am definitely gonna check out Prof. Francis channel on that, thanks for the reference. Dr.Bob, just my two cents here, but don't you think its better to make more math videos in areas which do not have good series of them available on youtube already, for example I am taking complex variables right now, and haven't found too much on youtube to refer too. For things like pre-calc, many people like khanacademy have great videos on that already.
I like you a lot, guy. You're doin' it right. Keep it up.
hey bob...hiw to find the number of homomorphisms in general between two groups? ex: z/21z to z/41z
+prudhvi raj Big question. For your example, it's greatly simplified since Z/21 is cyclic. Then any homomorphism is determined by sending a generator (say 1) to an element with order that divides 21. So only the trivial homomorphism (all to 0) for Z/41, but interesting if Z/3 or Z/15, for examples.
I see that you have used two different kinds of arrows to denote a homomorphism from G to H. One uses a double-headed arrow (both heads facing the same way)and the other uses a single-headed arrow. Is there a different meaning to each of these?
Double headed arrow = surjection; hook arrow = injection
Hi Bob, i have a question on the example at 8:15, Z6 ~ Z2 x Z3 . I am confused on how we can determine if they are isomorphic and how did u get the order of each element in Z2 x Z3. Thanks so much!!~~
Hey, I have a question, in the theorem, the fact that ker (pi)=N is part of the hypothesis or the thesis?. Thanks in advance.
I understand, but think of it this way: The Galois theory videos I'm doing right now are a lot of fun, but they can be a brutal slog because it is very easy to make errors. If I did only middle to upper undergraduate material, I'd burn out in no time.
thanks a lot doctor Bob. you answered my question. thanks!!
a question please MathDoctorBob, Decide if the following claim is true or false, and give arguments to support your answer. The claim is that any two abelian groups of order 99 are isomorphic. I think that they are isomorphic since they are the same order and are both abelian please help.
In general, order is not enough to distinguish abelian groups. In the case of order 99, z/3 x z/33 and z/99 are not isomorphic. z/3 x z/33 doesn't have any elements of order 9 or 99 (only 1, 3, 11, and 33). The general result is the fundamental theorem of finite abelian groups.
Hi doctor bob! just wanna ask in the proof of Z6 is iso to Z2 x Z3 how does the proof of homomorphism go?
thanks for the wonderful videos! I'm learning algebra easily :)
how can i count the number of isomorphic classes for a 7 vertices each with degree 2 graph ?
There's a graph theory channel that can help with that one. Graph theory isn't my jurisdiction.
Did you mes up at 8.30. shouldn't pi(1)=(1,2)
That works also. There can be more than one isomorphism when they exist. 2 here since we need only send 1 to an element of order 6.
Your welcome!
Thanks! - Bob
thanks doctor bob :)... This is very helpful :D
Thanks! Someone has to, and until they come along, it's me. :)
Check out nptelhrd for a full complex analysis video course and other higher math.
I don't understand your teaching
+Jazmin Sanchez (_jazzyboo26) Teaching is not one size fits all! The videos for the Harvard course are a pretty good alternative.
+MathDoctorBob I know what you mean I have so much trouble understanding my professor's teaching. Thank you for the suggestion though!
yeah me neither it feels like he's just reading of a scrip or something, like a robot. I don't feel like he even understands it himself so it's really hard to understand