- #1

- 49

- 3

- Homework Statement
- see title

- Relevant Equations
- see below

Progress:

π:π(3)ββ€2

π:π(3)βππ(3)

π:π(3)/ππ(3)ββ€2

π(π)=det(π)

with πβπ(3), that way

π(π)β¦{β1,1}β β€2,

where 1 is the identity element.

Ker(π) = {πβππ(3)|π(π)=1}=ππ(3), since det(π)=1 for πβππ(3).

By the multiplicative property of the determinant function, π = homomorphism.

***What is the form of the canonical homomorphism (π) in this case?

I'm used to the coset language,

i.e. π:πΊβπΊ/Ker(π)

with π(π)=ππΎ for πΎ=ker(π)

If this were settled, then π is an isomorphism by the isomorphism theorem.

π:π(3)ββ€2

π:π(3)βππ(3)

π:π(3)/ππ(3)ββ€2

π(π)=det(π)

with πβπ(3), that way

π(π)β¦{β1,1}β β€2,

where 1 is the identity element.

Ker(π) = {πβππ(3)|π(π)=1}=ππ(3), since det(π)=1 for πβππ(3).

By the multiplicative property of the determinant function, π = homomorphism.

***What is the form of the canonical homomorphism (π) in this case?

I'm used to the coset language,

i.e. π:πΊβπΊ/Ker(π)

with π(π)=ππΎ for πΎ=ker(π)

If this were settled, then π is an isomorphism by the isomorphism theorem.