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Math Help - Homomorphism question.

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Homomorphism question.

    I think I'm confused about notation perhaps.

    let p: \mathbb{Z} x \mathbb{Z} -> G be a homomorphism

    define p(m,n)=h^{m}k^{n} for h,k in G

    give a necessary and sufficient condition for it to be a homomorphism... prove why.

    i know to show something is a homomorphism i have to show that p(ab)=p(a)p(b), ie. that the binary operation's action is preserved in both groups. but i don't know how to here. for a=(m,n) and b=(i,j), is ab=(mi,nj)? that would give h^{mi}k^{nj}=h^{m+i}k^{n+j}. How do I go about setting conditions to make these equal...? Seems to me they only would be if h,k were 0 or 1?
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    Quote Originally Posted by platinumpimp68plus1 View Post
    I think I'm confused about notation perhaps.

    let p: \mathbb{Z} x \mathbb{Z} -> G be a homomorphism

    define p(m,n)=h^{m}k^{n} for h,k in G

    give a necessary and sufficient condition for it to be a homomorphism... prove why.

    i know to show something is a homomorphism i have to show that p(ab)=p(a)p(b), ie. that the binary operation's action is preserved in both groups. but i don't know how to here. for a=(m,n) and b=(i,j), is ab=(mi,nj)? that would give h^{mi}k^{nj}=h^{m+i}k^{n+j}. How do I go about setting conditions to make these equal...? Seems to me they only would be if h,k were 0 or 1?
    I think it must be \mathbb{Z}\oplus\mathbb{Z} , which sometimes is simply written \mathbb{Z}\times\mathbb{Z} = the free abelian group of rank 2.

    The operation here's defined as (m,n)+(m',n')=(m+m',n+n') , so your function's a homom. iff p(m+m',n+n')=p(m,n)+p(m',n')\Longleftrightarrow\,\  ,h^{m+m'}k^{n+n'}=(h^m k^n)(h^{m'} k^{n'})

    Can you see now what the condition on G has to be?

    Tonio
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  3. #3
    Junior Member platinumpimp68plus1's Avatar
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    Ooh okay, I always get confused about notation.

    So G has to be abelian then, ie. hk=kh ?
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    Quote Originally Posted by platinumpimp68plus1 View Post
    Ooh okay, I always get confused about notation.

    So G has to be abelian then, ie. hk=kh ?

    It seems so, but the way you defined h doesn't make it clear: what are those two elements in G? Generators or what? Or h, k are only two arbitrary elements in G? If the latter then all it's requires is that h, k commute, though G itself can be non-abelian.

    Tonio
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  5. #5
    Junior Member platinumpimp68plus1's Avatar
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    Quote Originally Posted by tonio View Post
    It seems so, but the way you defined h doesn't make it clear: what are those two elements in G? Generators or what? Or h, k are only two arbitrary elements in G? If the latter then all it's requires is that h, k commute, though G itself can be non-abelian.

    Tonio

    they were just defined as arbitrary elements in G.

    thanks for your help.
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