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Math Help - Homomorphisms and kernels

  1. #1
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    Homomorphisms and kernels

    Let c:G1--->G2 and d:G2--->G3 be group homomorphisms. Prove that dc:G1--->G3 is a homomorphism. Prove that ker(c) is a subset of ker(d).

    Since c and d are homomorphisms, we know c(ab)=c(a)c(b) and d(ab)=d(a)d(b).
    We want to show dc(ab)=dc(a)dc(b)
    I get stuck at this point.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Let c:G1--->G2 and d:G2--->G3 be group homomorphisms. Prove that dc:G1--->G3 is a homomorphism. Prove that ker(c) is a subset of ker(d).

    Since c and d are homomorphisms, we know c(ab)=c(a)c(b) and d(ab)=d(a)d(b).
    We want to show dc(ab)=dc(a)dc(b)
    I get stuck at this point.
    I think you're making this harder than it is. If a,b\in G_1 then c(ab)=c(a)c(b)\in G_2 and so d(c(ab))=d(c(a)c(b))=d(c(a))d(c(b))\in G_3.

    I'll let you think about the second part.
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  3. #3
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    ker c is defined as x in G1 such that c(x)=e
    ker d is defined as x in G2 such that d(x)=e
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    ker c is defined as x in G1 such that c(x)=e
    ker d is defined as x in G2 such that d(x)=e
    Ok, so what does that tell us?
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  5. #5
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    since c(x) is in G2 then ker c is a subset of ker d
    I'm not sure
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  6. #6
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    Well, first, trying to say \mathrm{ker}c\subseteq \mathrm{ker}d doesn't make any sense; the two are not even contained in the same group.

    I believe the question should be to show that \mathrm{ker}c\subseteq \mathrm{ker}(dc)

    Now, just think about it; what would it mean to say \mathrm{ker}c\subseteq \mathrm{ker}(dc)?

    Purely by definition of what it means to be a subset, it means that we want to show that, if x\in \mathrm{ker}c, then x\in \mathrm{ker}(dc) as well. Why should this be true?
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  7. #7
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    You are right. Guess I misread the question.
    ker dc is defined as x in G1 such that dc(x)=e
    Since they both involve G1 it must be a subset?
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  8. #8
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    Is that anywhere near the right track? I guess I'm having trouble applying the definition.
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