Random post: Interesting quotient groups.

Warning: Pre-knowledge of group theory/Abstract Algebra is needed for this post.

Most oftenly when we think of quotient groups, we think of the modular arithmetic ones. But what about some other ones?

For example, ℝ/ℤ with respect to addition. Every element of ℝ/ℤ is by definition a coset of ℤ w.r.t ℝ. So every element can be described as ℤ+x, for some x∈[0,1) (cosets repeat themselves so we can restrict it this way). Now intuitively, we can think of this as a circle, cause when you have had one rotation; you repeat the whole process again. That's what we shall prove; (ℝ/ℤ,+)≌({e^(iθ)|θ∈ℝ},•). Let φ(ℤ+x)=e^(ix). Then it's obvious by exponentiation laws that this is a homomorphism. Now e^(ix)=e^(iy) => x-y is a multiple of 2π=> ℤ+x=ℤ+y. hence injectiveness. Also for every y=e^(ix) we can simply take x to see that ∃x∈ℝ φ(x)=y, hence f is surjective. In other words, the set is already defined as φ(ℝ), so its already defined to be surjective. Therefore, φ is bijective and thus indeed a isomorphism.

Another example, this time matrices. Let's consider the quotientgroup GL₂(ℚ)/SL₂(ℚ). Then any element of this quotientgroup is of the form A•SL₂(ℚ), ∃A∈GL₂(ℚ). Then we theorize that GL₂(ℚ)/SL₂(ℚ)≌ℚ\{0}. The function we'll prove to be isomorphism for this is the determinant; in the way that φ(A•SL₂(ℚ))=det(A). We firstly have to show that it's even well-defined; in the sense that if A•SL₂(ℚ)=B•SL₂(ℚ) implies det(A)=det(B). Consider C,D∈SL₂(ℚ) s.t. AC=BD, which exists by definition of our previous equation. Then det(A)det(C)=det(B)det(D)=>det(A)•1=det(B)•1=>det(A)=det(B). Therefor this is well defined. Then this is a homomorphism because det(AB)=det(A)det(B), and the definition of operations on quotient groups makes this complete. To show that its bijective; if det(A)=det(B) then A(A⁻¹BC)=BC∈A•SL₂(ℚ) since det(A⁻¹BC)=det(C). this works for any C, so A•SL₂(ℚ)=B•SL₂(ℚ). Next for subjectiveness, just pick aI₂SL₂(ℚ) for any a∈ℚ\{0}, and you see that aI₂∈GL₂(ℚ) and det(aI₂)=a so any a∈ℚ\{0} can be reached. Hence φ is bijective too, therefor also an isomorphism. Oh btw I just realized I picked 2 dimensions but there's no reason for that.

Thanks for reading