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Abstract Algebra (Normal Subgroups)

  1. Nov 1, 2005 #1
    Hello. We got a review today in abstract algebra, and I am stuck on two problems.

    1) Let f: G -> H be a surjective homomorphism of groups. Prove that if K is a normal subgroup of G, then f(K) is a normal subgroup of H. Where f(K)= {f(k): k [tex]\in[/tex]K}
    The entire f(K) part is really throwing me off.
    We know:
    G and H are groups
    K is a normal subgroup of G
    Also H is a normal subgroup because im_f = H (because f is surjective)
    The real problem is I don't quite understand the whole f(K) part. Any ideas about what f(K) is, and any ideas about the problem?




    2) If H and K are normal subgroups of a group G satisfying H[tex]\cap[/tex]K = {1}, prove that hk = kh for all h[tex]\in[/tex]H and k[tex]\in[/tex]K.
    This one is really throwing me off.
    We Know:
    H and K are normal subgroups of a group G.
    H[tex]\cap[/tex]K = {1} is a normal subgroup of G. (because of a previous problem I had proved).
    Also, HK is a subgroup (by a theorem from my book)
    For this problem it seems that it would be sufficient to show that G is abelian, but I am not sure how we would do that. Or maybe just use some general properties algebraically to show that hk=kh. I am not sure how we are supposed to use the H[tex]\cap[/tex]K = {1} though.
    Any ideas about this one?




    Thanks!!!
     
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2
    subgroup test comes to mind. oh wait never mind you have to prove normal subgroup sorry prolly not that easy then
     
    Last edited: Nov 1, 2005
  4. Nov 1, 2005 #3
    Yes, but I am not understanding how I would say that 1 is in f(K), or x,y in f(K) => xy in f(K), my real problem is understanding how I would use f(K) to see if the properties of a subgroup apply to f(K). Any ideas about f(K)?
     
  5. Nov 1, 2005 #4
    yes I had a problem like this but it was dealing with isomorphisms I think I assumed f(a) and f(b) were in f(k) then I said f(a)f(b) was in f(k) since ab is in k so f(ab) is in f(k) and f(ab)=f(a)f(b), similar argument holds for a inverse

    f(k) is non empty because f(e) is in f(k)

    since k is a group it has all the properties we need like identity and a,b in k implies ab in k so f(ab) is in f(k)
    a in k then a^-1 in k so f(a^-1) is f(k)

    f(a^-1) = (f(a))^-1
     
    Last edited: Nov 1, 2005
  6. Nov 1, 2005 #5
    Hmm, that is an interesting idea, I think I remember my teacher doing something similar, I will try it out, thanks. Any ideas for #2?
     
  7. Nov 2, 2005 #6
    for the 1st one you've got to show that the subgroup f(K) is normal in H, that is, [tex]f(K) = f(x)f(K)f(x)^{-1}[/tex] for any f(x) in H if K is normal in G.
    unless i made a mistake (or i'm getting better at algebra) it's not very hard. all you've got to use, after all, is properties of homomorphisms to show that the image is a normal subgroup. i think the reason you need a surjective homomorphism is so that you get a group in the image, otherwise you could get the empty set as an image which isn't a normal subgroup.
    for the 2nd one i'd try out some examples. the quaternions is a noncommutative group in which every subgroup is normal, so you're idea won't work (showing it's abelian). try playing with a concrete example like the quaternions & see if that helps.
     
  8. Nov 2, 2005 #7

    AKG

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    Suppose x is in f(K) and h is in H. To show f(K) is normal, you want to show that hxh-1 is in f(K). Well you know that since f is surjective, there is some g in G such that f(g) = h. Since x is in f(K), you know there is some k in K such that f(k) = x. So you can pick g and k such that:

    hxh-1 = ... (figure the rest out)
    Do you know that if H is normal, then for every g in G, gH = Hg? In other words, for every g in G and for every h in H, there is some h' in H (possibly the same as h) such that gh = h'g. This holds for every g in G, including every g in K. Can you go from here?
     
  9. Nov 2, 2005 #8
    yeah i think i made a mistake. i should think harder
     
  10. Nov 2, 2005 #9
    Thanks everyone, I got them both :smile:
     
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