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Showing that a certain subgroup is the kernal of a homomorphism

  1. Oct 4, 2011 #1
    Let Q be the quaternion group {1, -1, i, -i, j, -j, k, -k}. Show that the normal subgroup {1, -1, i, -i} is the kernal of a homomorphism from Q to {1, -1}.

    I know that if N is a normal subgroup of G then the homomorphism f: G -> G/N has N as the kernal of f. while i can get the kernal of f to be {1, -1, i, -i} i can't seem to get the codomain to be {1, -1}. i've gone through different mappings and all of them while they had the correct kernal always seem to have more than just {1, -1} in the codomain. can someone help me determine the correct homomorphism?
     
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  3. Oct 5, 2011 #2

    Hurkyl

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    Can you show your work? Why do you think the codomain has more than 2 elements?

    Every element of Q appears in the image. The point is that they aren't all distinct in the image.
     
  4. Oct 7, 2011 #3
    i have tried mapping x to x^2 but then i2 = -1 so it is not in the kernal, x^4 would exclude -1 in the set {1, -1}. i have also tried mapping x to xN where N = {1, -1, i, -i} is a normal subgroup so there exists a natural homomorphism with N as its kernal. but that doesn't work since for instance when x = k we would get kN = {k, -k, -j, j} which is not the right codomain.

    would this next map work? [itex] x \mapsto xN [/itex] where N = {1, -1, i. -i}, x is any member of Q, xN is a left coset, and N is a normal subgroup of Q. Then i map again: [itex] xN \mapsto (xN)^2 [/itex]. so the composition of these will map an x to (xN)2 and the codomain will be {1, -1}.
     
  5. Oct 7, 2011 #4

    Deveno

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    no because the map x→(xN)2 maps everything to the trivial group {N}, so its kernel is Q.

    you were on the right track mapping x→xN, can you think of a way to create an isomorphism of Q/N with {-1,1}?
     
  6. Oct 10, 2011 #5
    ok so i know i can't square it since i would just get x2N = N which would be the identity. i was thinking earlier [tex] xN \mapsto x(xN) [/tex] but that just gives me N back again. what i want to do is to "kinda" square all the entries somehow in xN so that i get only 1 and -1 but i cannot figure out a mapping to help me do that.
     
  7. Oct 11, 2011 #6
    how about [itex] xN \mapsto x^2 [/itex]? that way {1, -1, i, -i} = N maps to 1 and {j, -j, k, -k} = (k)^2 = -1
     
  8. Oct 11, 2011 #7

    Deveno

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    the map xN → x2 is not well-defined.

    for x = i, we get N→i2 = -1
    for x = 1. we get N→12 = 1

    so we do not have a unique image for the coset N.
     
  9. Oct 13, 2011 #8
    hm i'm really stumped on this. it's probably not that difficult but i just don't see it. can someone offer a hint or two?
     
  10. Oct 14, 2011 #9

    Deveno

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    re-read post #4.

    if N = {1,-1,i,-i} then φ:Q→Q/N given by x→xN, has kernel N.

    since Q/N and {-1,1} are both cyclic groups of order 2, there is an isomorphism between them, namely:

    N→1
    jN(= kN)→-1, if we call this isomorphism ψ, then ψφ is the desired homomorphism.
     
  11. Oct 14, 2011 #10
    ah i didn't notice that they were cyclic groups! i was trying to come up with an explicit mapping between them (like a formula of some sort) but could not come up with one at all. this is my first time being exposed to these kind of ideas and i think i'm slowly starting to understand more of it. thanks for your help!
     
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