What is the kernel of the determinant mapping in GL(2,R)?

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Homework Help Overview

The discussion revolves around the kernel of the determinant mapping in the context of the groups GL(2,R) and SL(2,R). Participants are exploring the properties of these groups and the nature of homomorphisms in relation to determinants.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are attempting to clarify the definitions of the kernel and the identity element in group theory, particularly in relation to the determinant mapping. Questions are raised about the nature of GL(2,R) as a group and the properties of homomorphisms.

Discussion Status

The discussion is active, with participants questioning and clarifying the definitions and properties of the groups involved. Some guidance has been offered regarding the identity element in R* and the nature of the kernel in the context of group homomorphisms.

Contextual Notes

There is some confusion regarding the terminology used for identity elements in groups versus rings, as well as the implications of the kernel in this specific mathematical context.

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Homework Statement




Let R* be the group of nonzero real numbersunder multiplications. Then the determinant mapping A->det A is a homomorphism from GL(2,R) to R* . The kernel of the determinant mapping is SL(2,R).

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The Attempt at a Solution



I know det(A)det(B)=det(AB) but other than knowing that property, I don't understand the meaning of the kernel nor SL(2,R) nor do I understand how GL(2,R) is a homomorphism. I know SL(2,R) stands for Special linear group and GL(2,R) General Linear group.
 
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GL(2,R) isn't a homomorphism. It is a group. Your post clearly states that the determinant map is a homomorphism, not GL(n,R).

What is det(I), I the identity?

Doesn't this show that det satisfies the definition of homomorphism?

You do understand what SL(2,R) is - you wrote out its definition: the set of matrices of determinant 1.

The kernel is the set of matrices sent to the identity...
 
Isn't the kernel the set of stuff that is sent to 0, not the identity?
 
theperthvan said:
Isn't the kernel the set of stuff that is sent to 0, not the identity?
Groups aren't even required to have an element called '0'!
 
Yeah true, so the definition I had must've been for something with identity=0. So is it really what is sent to the identity?
 
In a group, the identity is often denoted e. The zero is the additive identity element in a ring. The wikipedia article calls the identity element (in a group) 1.

So, calling the additive identity either 0 or 1 is generally a bad idea. (the '1' is actually 0 in the group of integers and all of its subgroups, and calling it '0' is...a ring thing)

The kernel of a group homomorphism phi:A->B is the preimage of {e_B} under phi, e_B the identity element in e_B. The preimage of a subset S of B under a function f:A->B is defined set theoretically as {x in A : f(x) in S}.

So for your problem: What's the identity element in R*? (Certainly not zero!) What's the preimage of this identity under the group homomorphism given by the determinant?

These should all be obvious to you.

(Note: SL(n,R) is defined as the nxn matrices over R with determinant 1. Exercise: Show that this is a subgroup of GL(n,R))
 
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