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Math Help - proving surjective homomorphism

  1. #1
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    proving surjective homomorphism

    Requesting a bit of help on the last part of my homework.

    Need to prove that the following is a surjective homomorphism:

    Let H and K be groups. Define the projection map:

    : H x K H such that : (h, k) h

    : H x K K such that : (h, k) k

    How do you prove something is onto?

    I know for a homomorphism you show that:
    (ab) = (a)(b)

    How would I attack this proof?

    Please help!

    Thanks.
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  2. #2
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    Quote Originally Posted by kpizle View Post
    Requesting a bit of help on the last part of my homework.

    Need to prove that the following is a surjective homomorphism:

    Let H and K be groups. Define the projection map:

    : H x K H such that : (h, k) h

    : H x K K such that : (h, k) k

    How do you prove something is onto?

    I know for a homomorphism you show that:
    (ab) = (a)(b)

    How would I attack this proof?

    Please help!

    Thanks.

    I'm guessing you must prove \pi_1\,,\,\pi_2 are surjective. Take the first one: since H\times K=\{(h,k)\,;\,\,h\in H\,,\,k\in K\}, then for ANY h\in H\,,\,\pi_i(h,1)=h and we have surjectivity.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    I'm guessing you must prove \pi_1\,,\,\pi_2 are surjective. Take the first one: since H\times K=\{(h,k)\,;\,\,h\in H\,,\,k\in K\}, then for ANY h\in H\,,\,\pi_i(h,1)=h and we have surjectivity.

    Tonio
    Ok.

    So,

    Let <br />
\,h, h', e\in H\,,\,k, k', e'\in K\

    (hh',k) = hh' (def )
    = h x h'
    = (h,k) x (h',k) (def operation H x K)

    Since (hh',k) = (h,k) x (h',k) is a homomorphism

    Since a homomorphism let \,e\in H\,,\,e'\in K\ be the identities.

    (h,e) = h (def )

    Therefore, a surjective homomorphism.

    By a similar argument, \pi_2 is also a surjective homomorphism.

    QED

    This feels very wrong to me...but rarely do proofs ever feel right.

    Am I on the right track?
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  4. #4
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    Quote Originally Posted by kpizle View Post
    Ok.

    So,

    Let <br />
\,h, h', e\in H\,,\,k, k', e'\in K\

    (hh',k) = hh' (def )
    = h x h'
    = (h,k) x (h',k) (def operation H x K)

    Since (hh',k) = (h,k) x (h',k) is a homomorphism

    Since a homomorphism let \,e\in H\,,\,e'\in K\ be the identities.

    (h,e) = h (def )

    Therefore, a surjective homomorphism.

    By a similar argument, \pi_2 is also a surjective homomorphism.

    QED

    This feels very wrong to me...but rarely do proofs ever feel right.

    Am I on the right track?

    \pi_1\left((h_1,k_1)(h_2,k_2)\right)=\pi_1(h_1h_2,  k_1k_2):=h_1h_2=\pi_1(h_1,k_1)\pi_1(h_2,k_2)

    This proves the maps are homomorphisms,a nd you already have surjectivity, so you're done.

    Tonio
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  5. #5
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    Ooohhh. The missing link!

    Thank you very much.
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