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

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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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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