If we are talking about homomorphism between two groups f:G \to G' for example, then only one condition would hold not two since a group is equipped with only one binary operation such that the group operation on the left side of the condition is that of G and that on the right hand side is that of G'. However, for ring homomorphism there are two conditions to satisfy.
@rhonkie I have checked my notes and you are perfectly right. This is mainly just a groups video on groups. Thanks for that...this was my worst topic and maybe it shows lol. thanks
yes the group is closed just under a singular binary operation. The ring on the other hand has a dual binary operation closure property. Homomorphism is analogous to linear transformations in matrices with the property that a homomorphism of two elements in a group maps to the sums of the * operation of the other groups.
I am not entirely clear about your definition of homomorphism. There are different kinds of homomorphism such as group homomorphism, ring homomorphism and so on. The one described at the start of your video is a little confusing to me.
@jaryH3 Ahh yes i see thats true. I think i used this from notes in lectures so i not quite sure how to fix it. Maybe i should have defined theta as an element of the natural numbers.(does that sound right to you?) But thanks for pointing that out.
@burny1 I guess that theta being a function N->N would help, but then, as it appears to me, the absolute value looses it's purpose.
Reading a Wikipedia (Which means that I don't know what I'm talking about.) I find you may be missing the whole point of homomorphism. The idea of h. may be that you have two groups G and G'. These are defined under different operation * and *' . You may, but not necessary, be able to find a function phi G->G' so that phi(a * b) = phi(a) *' phi(b).
@burny1 ... and it preserves the structure. Instead of doing the operation in * on a and b in G, you may map them to a'=phi(a) and b'=phi(b), do the a' *' b' and then map it back using inverse_phi:
If we are talking about homomorphism between two groups f:G \to G' for example, then only one condition would hold not two since a group is equipped with only one binary operation such that the group operation on the left side of the condition is that of G and that on the right hand side is that of G'. However, for ring homomorphism there are two conditions to satisfy.
rhonkie 1 year ago
@rhonkie I have checked my notes and you are perfectly right. This is mainly just a groups video on groups. Thanks for that...this was my worst topic and maybe it shows lol. thanks
burny1 1 year ago
@rhonkie
yes the group is closed just under a singular binary operation. The ring on the other hand has a dual binary operation closure property. Homomorphism is analogous to linear transformations in matrices with the property that a homomorphism of two elements in a group maps to the sums of the * operation of the other groups.
17thsavior 1 year ago
I am not entirely clear about your definition of homomorphism. There are different kinds of homomorphism such as group homomorphism, ring homomorphism and so on. The one described at the start of your video is a little confusing to me.
rhonkie 1 year ago
Hey! The statement
|a+b| = |a| + |b|
is not correct for a,b \in Real. In fact:
|a+b| <= |a| + |b|".
Take "a=-3" and "b=2" for example. Definitely:
|-3+2| = 1 < 5 = |-3|+|2|".
Sorry for the formatting. YT seems to not work well with math. Also sorry if I misunderstood the group theory.
jaryH3 1 year ago 2
@jaryH3 Ahh yes i see thats true. I think i used this from notes in lectures so i not quite sure how to fix it. Maybe i should have defined theta as an element of the natural numbers.(does that sound right to you?) But thanks for pointing that out.
burny1 1 year ago
@burny1 I guess that theta being a function N->N would help, but then, as it appears to me, the absolute value looses it's purpose.
Reading a Wikipedia (Which means that I don't know what I'm talking about.) I find you may be missing the whole point of homomorphism. The idea of h. may be that you have two groups G and G'. These are defined under different operation * and *' . You may, but not necessary, be able to find a function phi G->G' so that phi(a * b) = phi(a) *' phi(b).
jaryH3 1 year ago
@burny1 ... and it preserves the structure. Instead of doing the operation in * on a and b in G, you may map them to a'=phi(a) and b'=phi(b), do the a' *' b' and then map it back using inverse_phi:
a * b = inverse_phi(phi(a) *' phi(b)).
Once again... I don't know whether I'm correct.
jaryH3 1 year ago
@burny1 Studying some of my materials I found:
Grupoid is a set closed by some operation.
If (G, ☘) and (H, ♡) are grupoids and w:G→H is a mapping, while
w(g1☘ g2) = w(g1) ♡ w(g2) g1, g2 ∈ G
then w is a homomorphism of grupoid (G, ☘) to grupoid (H, ♡).
Additionaly, if w is bijective, then w is an isomorphism. And even more, if G=H, then w is an automorphism.
Hope it helps someone.
jaryH3 1 year ago