First Isomorphism Theorem of Groups -- Abstract Algebra Examples 14

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@24:17 why are you setting phi(x)=x
For the kernel, shouldn’t we be setting it to the identity e?

Happy_Abe

At [20:01], he conflates elements of G with elements of Inn(G). The latter are *functions* operating on G(!).

His comment that "There is a lot that is inherited from the parent group" is not correct.

GlenMacDonald

Typo at 7:30, it should be "phi(n) ≡ 0 mod 15", but the next line is correct - this typo causes Justin to say "domain" instead of "codomain" at 8:33

Also at 27:55 it is obvious that exp(x) = exp(y) implies x = y, but that's not a _proof_ that exp is injective because this is just saying "exp is injective because exp is injective" - of course what you are implicitly doing is using the logarithm. Indeed another (more "categorical") way to show that exp is an isomorphism from *R* to *R* ^{×}_{+} is to show that the inverse function log: *R* ^{×}_{+} → *R* is also a homomorphism, which is the log property log(xy) = log(x) + log(y).

If you so desired, you can show that exp is injective without using its inverse function, using any of the equivalent definitions of the exponential function. For example if you define exp by the power series exp(x) := Σ x^n/n! or the limit exp(x) := lim_{n→∞} (1 + x/n)^n then it is easy to show that exp is strictly increasing and thus injective. If instead you define exp as the unique solution to the differential equation exp' = exp and exp(0) = 1 then you can easily derive the power series form which we already know is injective. Finally one can define exp _by its homomorphism property_ : exp is the unique continuous function from *R* to *R* ^{×} = *R* - {0} such that exp(x+y) = exp(x)exp(y) and exp(1) = e (where the constant e has been defined in a different way) - annoyingly we have to include "continuous" (or "increasing" will work too) because otherwise the Axiom of Choice allows for pathological solutions to the functional equation f(x+y) = f(x)f(y), however once we specify that it is continuous then it is easy to show that it is differentiable and satisfies the differential equation exp' = exp and exp(0) = 1, and hence is injective.

The fact that the axiom of choice allows pathological functions solving f(x+y) = f(x)f(y) has the _algebraic_ consequence that the exponential functions may not be the only homomorphisms from *R* to *R* ^{×}, depending on the model of set theory - since *R* and *R* ^{×} are *Z* -modules (abelian groups), we get a *Z* -module of homomorphisms Hom_ *Z* ( *R*, *R* ^{×} ), and therefore whether the axiom of choice holds or not affects this module: in any model of ZFC we have that Hom_ *Z* ( *R*, *R* ^{×} ) is strictly bigger than the module { x ↦ a^x | a > 0 }, whereas there are models of ZF (necessarily models of ZF-C) in which Hom_ *Z* ( *R*, *R* ^{×} ) = { x ↦ a^x | a > 0 } ≅ *R* ^{×}_{+}

schweinmachtbree

@ 20:40 shouldn't you prove that there exists ψ^-1 such that ψ ∘ ψ^-1 = ψ^-1 ∘ ψ = id_G, which means ψ(ψ^-1(x)) = ψ^-1(ψ(x)) = x?
In this case we would have ψ^-1(x) = g^-1 x g, which is not (g x g^-1)^-1.

davidemasi__

1:40 Curiously, it doesn't require f to be an homomorphism. Any (total) function f would work.
You mostly rely on the fact that for all relevant x, f(x) exists and is unique; And the fact that equality is an equivalence relation.

matheusjahnke

Does Randolph have grad classes in either Galois or Representation theory?

robshaw

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