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Homomorphisms and kernals

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data

    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).
    2. Relevant equations

    3. 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.
  2. jcsd
  3. Oct 28, 2007 #2

    matt grime

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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...
  4. Oct 28, 2007 #3
    Isn't the kernel the set of stuff that is sent to 0, not the identity?
  5. Oct 28, 2007 #4


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    Groups aren't even required to have an element called '0'!
  6. Nov 13, 2007 #5
    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?
  7. Nov 13, 2007 #6
    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))
    Last edited: Nov 13, 2007
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