Basic abstract algebra, pt.5

This will be a video series about basic abstract algebra and group theory. We will keep it basic so that anyone can follow the videoes.

This is the fifth video in the series, and we will define a tool to compare groups. We define homomorphisms and isomorphisms between groups.

Category:

Education

Tags:

• basic
• abstract
• algebra
• group
• theory
• mathematics
• math
• theorem
• prove
• proof
• homomorphism
• isomorphism
• bijection
• science

[BruceLCM]

sorry, i mean bijective instead of one-to-one

[VeritySeeker]

@BruceLCM Not all functions are bijective, and not all functions are injective, nor surjective. It is your choosing. But if the sets have the same cardinality, a bijection is possible by definition. I think we are on the same page :).

[voluptuate]

Hi, a comment. You state that if two groups are the same then the same binary operator should behave the same way in both groups. You gave a formula expressing this as f (a*b) = f(a)*(b). I can't see why if your intent is to see if operator behaves the same you would do a different thing between the groups. Wouldn't you want to test if f (a*b) in group F = f (A*B) in group G? Now you will see if the same operation gives the same result in two groups?

[VeritySeeker]

Hi, this is a good question. I have to ask, though, what are the elements A and B in your question? I think I know what you mean, and I will make an attempt to explain. If I misunderstood you, just ask again.

Ok, here it goes: I have two groups G and G', and a homomorphism f from G to G'. The function f is a correspondence between the groups, so if a is in G, then f(a) is an element in G'. I want to show that f(a)*f(b), which are an element in G' corresponds to a*b in G VIA the function f.

[VeritySeeker]

That would mean that f(a*b) = f(a)*f(b), where the first * is the operation in G and the second * is the operation in G'. In your question, however, the operation * is in the group G, and it doesn't involve the operation in G' at all. So what I really want to test is:

[VeritySeeker]

Take two elements in G, say a and b and form the element a*b. Now I want to check that the corresponding element to a, namely f(a) multiplied with the corresponding element of b, namely f(b) is the element corresponding to a*b, namely f(a*b). In other words: f(a*b) = f(a)*f(b).

f sends an element from G to G'. But in your example, the elements f(a*b) and F=f(A*B) are bot elements in G'. a,b,A and B are all elements in G for it to make any sense.

I hope this was clear. If not, just ask.

[zxtonizx]

Thank you for taking the time out to make these videos. They tighten up my basic understanding of the concepts. I'm studying finite albelian groups right now and it''s easy to get stuck under all the language. This brings me back on top. Thanks again

[pyrofreakingmaniac]

Up till this video, I considered myself more of a calculus/analytical/geometry kind of person, but ... this is just beautiful. Isomorphisms-- wow. It's a whole new way of thinking that opens up so many beautiful new possibilities!

[Kurtlane]

Part 4 was blocked because of some copyright EMI content violation. Please remove it and reload asap.

[Yaldatchayil]

I love this series. I'm a music student studying mathematics and music theory, and I want to beef up my math background.

To clarify:

a "binary operation" is the operation that governs a group G, correct? That is, the operation * such that a*b = c, where a, b and c are elements of G?

Also, I would really like to see Part 4, but it's been blocked by EMI "due to copyright issues." Could you please repost it sans song?

[Tarkvinius]

oh, EMI has blocked it. looks like the song you used was copyrighted. could you repost it without the song, please?

[Tarkvinius]

what happened to video 4?

[screamformealan]

Why can't we view video #4

[BruceLCM]

The video said it is always "possible" to make a one-to-one function when the domain and range have the same number of elements but not "necessarily" all functions are one-to-one.

[BruceLCM]

I thought the same before. What u said is so true! Not all the elements in the range of a function have a corresponding value from the domain. In the example given, A and B have the same number of elements. Yes, if we assign them to a function randomly, f: A->B may not be one-to-one, but we can deliberately make it! Say if A={1,2,3,4};B={5,6,7,8}. I can make f(1) = 5, f(2) = 8, f(3) = 7, f(4) = 6. Now, B is entirely covered under f: A->B.

[SirBastian200]

...but to me, this doesn't preclude the possibility of extra elements hanging around in A that don't get mapped to B at all.

My guess here is that the terminology f: A -> B implicitly maps all of A, but doesn't necessarily cover B. So if we have some function:

f: Integers between 4 and 1089 -> Users on youtube

then we expect f to have a value for any such integer, but we needn't expect the image of f to contain every user on youtube. Am I right?