Visual Group Theory, Lecture 4.3: The fundamental homomorphism theorem

Visual Group Theory, Lecture 4.3: The fundamental homomorphism theorem
The fundamental homomorphism theorem (FHT), also called the "first isomorphism theorem", says that the quotient of a domain by the kernel of a homomorphism is isomorphic to the image. We motivate this with Cayley diagrams before formally stating and proving it. This gives us a 2nd way to prove that two groups are isomorphic, which is often easier than constructing an explicit isomorphism. We conclude by applying the FHT to construct cyclic groups as natural quotients of the integers.
Course webpage (with lecture notes, HW, etc.): www.math.clemso...

[@kevinbyrne4538]

2:38 -- I appreciate the insight that a homomorphism is an isomorphism between cosets in the domain and elements in the image.

[@sahhaf1234]

V_4 is the klein 4-group, for people who have forgotten. According to wiki, this is equivalent to the dihedral group of order 2, D_2.

[@maxbow-arrow5931]

23:00 Are you sure there's no mistake? Shouldn't it be positive natural numbers? Maybe I'm mistaken, but I got stuck on this part for an hour or so, to realize that: Given a morphism f: Q*->Q+; f(xy)=f(x)+f(y); f(1)=0; Ker(f) = Let f(2) = a/b, f(3) = c/d Then f(2^bc) = bc*f(2) = ac = ad*f(3) = f(3^ad) Since the kernel should be {-1; 1}, it has to be either 2^bc=3^ad or 2^bc=-3^ad. Since both 2^bc and 3^ad are positive, second scenario is impossible; and the first one is only possible when bc=ad=0. Since b=0 or d=0 would make f(2) or f(3) out of codomain, we see that a=c=0. But then f(2)=f(3) and that expands the kernel to include both of them, which is a contradiction.

[@rasraster]

Couple of errors in the last example. Isomorphism is from Z/ --> Z12. Also, your i is actually the same.

[@partialorder5596]

At 23:06, you mean Q*/ is isomorphic to the positive rationals under multiplication, right?

[@kikones34]

Yes, that's what the notation actually means, he just misspoke.

[@quico522]

greetings from spain!

[@nktcp6908]

15:25, do you mean for all g in G, not for all h in G?

[@hyperduality2838]

Injective is dual to surjective synthesizes bijective or isomorphism. Similarity or equivalence implies duality! Domains are dual to co-domains --> homomorphisms. Points are dual to lines -- the principle of duality in geometry. "Always two there are" -- Yoda.

[@rasraster]

Is it OK to assume that phi IS a homomorphism when determining whether it's well-defined, BEFORE determining that it actually is a homomorphism? Seems like a catch-22: it's not a homomorphism if it's not well defined, but it can't be well defined if it's not a homomorphism...

[@rasraster]

My mistake. Phi is assumed to be a homomorphism; you are showing that i is also.

[@rasraster]

The argument about i being well-defined uncovers a really interesting fact: if you take any element of a left coset and left-multiply the coset by that element's inverse, that takes you back to the original subgroup of the coset! Which makes sense when you think about how left cosets just translate the subgroup to elsewhere in the overall group. Cool!

[@rasraster]

And taking things further, this ability to use the inverse of any representative is manifested in the product of some coset element and the inverse of any coset element being made essentially equivalent. That gets borne out by the morphism with the image of the kernel (the identity). So for the purposes of mapping to the identity, any representative is as good as any other.