All these videos on group theory are really very helping. After following them now I have crystal clear understanding of various facts. Thanks a lot for sharing
I like the ending. Because throughout the homomorphism series, I kept asking myself what's the point of all those equations. Seemed like a pointless game that some bored mathematicians invented. But the ending does justice to the series. Thank you.
Anyone viewing a video about group theory probably already know this but if it isn't obvious that the sum of two even numbers are even consider that an even integer is a multiple of 2, so it is of the form 2*q. An odd number is always "one plus" an even number, so it's of the form 2*q + 1. When adding two even numbers we get (2*q) + (2*Q) = 2(q + Q), i.e. the sum is also even. If we add two odd numbers we get (2*q + 1) + (2*Q + 1) = 2(q + Q + 1), so it's also always even. However adding an odd and an even integer gives: (2q) + (2*Q + 1) = 2q + 2Q + 1 = 2(q + Q) + 1, i.e. an even number "plus one" which therefore is an odd number.
Yeah, but if we have two groups (one being the domain, and another being the codomain), how can we come up with a mapping that will work as a homomorphism? That is, without the need of going through all possible elements and trying them out in this homomorphism law, because for larger groups this would be a tedious task. Even worse, for INFINITE groups, it might actually be impossible to check them all :P So how can we come up with a mapping that will work for sure? Provided that there in fact is such a mapping, that is. And if there isn't, how can we tell that without having to try all possible mappings? (Because that would be even worse than checking just one.)
Excellent question. If you happen to know a presentation of the domain group, then you can reduce your work significantly. What is a presentation? A presentation is a way of specifying the generators of a group and relations among the generators. So, for example, to describe the cyclic group of order 5, you can say C_5 = where e is assumed to be the identity element. The left hand side of the colon is the generator(s). In this case, there is only one generator. On the right hand side, we have the relation(s). As an example of a noncyclic group D_8 (or D_4 if you prefer to name your dihedral groups after the number of sides rather than the order) = It's hard to determine group presentations. But you can know you have a full group presentation if you have a relation which gives an order to each generator and explains how all pairs of generators "interchange" with each other. If you have that, then you definitely can fully explain the group multiplication. (Technically, you don't need to specify an order for every generator, but not doing so gives that generator infinite order. For example, would be the cyclic group of infinite order. Similarly, you don't have to specify ways to interchange generators, but doing so means they can't be interchanged at all. So if you had , then you still get infinitely many elements since there is no way to rewrite any of the following: a, ab, aba, abab, ababa, ababab, etc.) Anyway, if you know a presentation of G, you can uniquely define a group homomorphism _out of_ G by specifying where the generators go and "extending the function multiplicatively". For example, if you know G = , then to define a homomorphism φ from G to any other group H, you define φ(a) and φ(b). Then, you just assume φ(a^2) = φ(a)^2 and φ(ab) = φ(a)φ(b). It's sort of a "homomorphism by definition". We do this with vector spaces too. Specifying where a basis is sent uniquely determines a linear transformation. *_However,_* we must be _more careful_ here with groups than we are with vector spaces. When defining the outputs of the generators of G, you _must make sure_ that the outputs of the generators of G satisfy the same relations that the generators of G satisfied, otherwise you will have an ill-defined function. And this is because elements of G can be written in many ways, so you need to know that every different way to write the same element is mapped to the same output. So in our example with G = , we only have to specify φ(a) and φ(b), but we also need to check that, in H, we have φ(a)^4 = e, φ(b)^6 = e, φ(a)φ(b) = φ(b)φ(a) If we pick elements of H so that the above are all true, then we have defined a valid group homomorphism from G to H. And conversely, if we have a valid group homomorphism from G to H, then all of the above will be true. To see this, suppose we pick φ(a) and φ(b) so that φ(a)φ(b) ≠ φ(b)φ(a). By the assumed multiplicativity of φ, we have φ(ab) ≠ φ(ba). However, ab = ba, so φ is ill-defined. Similarly, if φ is a well-defined homomorphism, then φ(ab) = φ(ba), so φ(a)φ(b) = φ(b)φ(a). So this greatly reduces the task of checking whether something is a homomorphism. Of course, this requires you to know a presentation of the domain group, which could be quite tricky.
All these videos on group theory are really very helping. After following them now I have crystal clear understanding of various facts. Thanks a lot for sharing
I like the ending. Because throughout the homomorphism series, I kept asking myself what's the point of all those equations. Seemed like a pointless game that some bored mathematicians invented. But the ending does justice to the series. Thank you.
You sir are a rebel. Nicely done.
Anyone viewing a video about group theory probably already know this but if it isn't obvious that the sum of two even numbers are even consider that an even integer is a multiple of 2, so it is of the form 2*q. An odd number is always "one plus" an even number, so it's of the form 2*q + 1.
When adding two even numbers we get (2*q) + (2*Q) = 2(q + Q), i.e. the sum is also even.
If we add two odd numbers we get (2*q + 1) + (2*Q + 1) = 2(q + Q + 1), so it's also always even.
However adding an odd and an even integer gives: (2q) + (2*Q + 1) = 2q + 2Q + 1 = 2(q + Q) + 1, i.e. an even number "plus one" which therefore is an odd number.
thank you!!!!! your videos have been EXTREMELY helpful for me, in aiding me with my modern algebra course.
Thank you for teaching us.
i really really thank you it is a lot to me
sir , in this example you have skipped zero,i have small confusion in it.
I am finding this videos very useful, thank you so much
Zero is an even number.
Zero is also the identity element in {Z, +} so it must map to the identity in the other group.
Yeah, but if we have two groups (one being the domain, and another being the codomain), how can we come up with a mapping that will work as a homomorphism? That is, without the need of going through all possible elements and trying them out in this homomorphism law, because for larger groups this would be a tedious task. Even worse, for INFINITE groups, it might actually be impossible to check them all :P So how can we come up with a mapping that will work for sure? Provided that there in fact is such a mapping, that is. And if there isn't, how can we tell that without having to try all possible mappings? (Because that would be even worse than checking just one.)
Excellent question. If you happen to know a presentation of the domain group, then you can reduce your work significantly.
What is a presentation? A presentation is a way of specifying the generators of a group and relations among the generators. So, for example, to describe the cyclic group of order 5, you can say
C_5 =
where e is assumed to be the identity element. The left hand side of the colon is the generator(s). In this case, there is only one generator. On the right hand side, we have the relation(s). As an example of a noncyclic group D_8 (or D_4 if you prefer to name your dihedral groups after the number of sides rather than the order)
=
It's hard to determine group presentations. But you can know you have a full group presentation if you have a relation which gives an order to each generator and explains how all pairs of generators "interchange" with each other. If you have that, then you definitely can fully explain the group multiplication. (Technically, you don't need to specify an order for every generator, but not doing so gives that generator infinite order. For example, would be the cyclic group of infinite order. Similarly, you don't have to specify ways to interchange generators, but doing so means they can't be interchanged at all. So if you had , then you still get infinitely many elements since there is no way to rewrite any of the following: a, ab, aba, abab, ababa, ababab, etc.)
Anyway, if you know a presentation of G, you can uniquely define a group homomorphism _out of_ G by specifying where the generators go and "extending the function multiplicatively". For example, if you know
G = ,
then to define a homomorphism φ from G to any other group H, you define φ(a) and φ(b). Then, you just assume φ(a^2) = φ(a)^2 and φ(ab) = φ(a)φ(b). It's sort of a "homomorphism by definition". We do this with vector spaces too. Specifying where a basis is sent uniquely determines a linear transformation.
*_However,_* we must be _more careful_ here with groups than we are with vector spaces. When defining the outputs of the generators of G, you _must make sure_ that the outputs of the generators of G satisfy the same relations that the generators of G satisfied, otherwise you will have an ill-defined function. And this is because elements of G can be written in many ways, so you need to know that every different way to write the same element is mapped to the same output.
So in our example with G = , we only have to specify φ(a) and φ(b), but we also need to check that, in H, we have
φ(a)^4 = e,
φ(b)^6 = e,
φ(a)φ(b) = φ(b)φ(a)
If we pick elements of H so that the above are all true, then we have defined a valid group homomorphism from G to H. And conversely, if we have a valid group homomorphism from G to H, then all of the above will be true.
To see this, suppose we pick φ(a) and φ(b) so that φ(a)φ(b) ≠ φ(b)φ(a). By the assumed multiplicativity of φ, we have φ(ab) ≠ φ(ba). However, ab = ba, so φ is ill-defined. Similarly, if φ is a well-defined homomorphism, then φ(ab) = φ(ba), so φ(a)φ(b) = φ(b)φ(a).
So this greatly reduces the task of checking whether something is a homomorphism. Of course, this requires you to know a presentation of the domain group, which could be quite tricky.