1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Abstract Algebra HW

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]N \lhd G [/itex] and [itex]K \vartriangleleft G[/itex] and [itex]N \cap K = \{e\}[/itex]. Show that if
    [itex]n \in N [/itex]and [itex]k \in K[/itex], then [itex]nk = kn[/itex]. Hint: [itex]nk = kn[/itex] if and
    only if [itex]nkn^{-1}k^{-1} = e[/itex].

    2. Relevant equations
    These "relevant equations" were not provided with the problem I'm just putting them here to make my solution more clear.
    [itex]e=k_1^{-1}k_1[/itex]
    [itex]e=n_1^{-1}n_1[/itex]


    3. The attempt at a solution
    Let [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
    Then
    [itex](n_1)(k_1)(n_2)(k_2)=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k1k2)=n1NK=NK[/itex]
    But
    [itex](n_1)(k_1)(n_2)(k_2)=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=KNk_2=Kk_2N=KN[/itex] where we used the fact that in this case, [itex]k_2 \notin N[/itex].
    Therefore [itex]NK=KN[/itex] for all [itex]n\in N[/itex] and [itex]k\in K[/itex]

    This seems correct to me but I didn't use the hint and my usage of [itex]N \cap K = \{e\}[/itex] seems a little hand wavey.

    Please help.
     
  2. jcsd
  3. Nov 15, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I didn't follow your proof. I think you are confusing things by using two elements from each subgroup. You only need the elements given in the problem statement: [itex]n \in N[/itex] and [itex]k \in K[/itex]. Now consider the element [itex]nkn^{-1}k^{-1}[/itex]. The goal is to show that this equals [itex]e[/itex]. One promising way to do this would be to show that it is an element of [itex]N \cap K[/itex].
     
  4. Nov 15, 2012 #3
    OK, I'll try your method next since I would like to know how to use the hint since the teacher is obviously trying to convey something he deems important. However, I modified my original method slightly and I think it works now. If someone has the time please tell if it does or how it doesn't work.

    Let [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
    Then
    [itex]{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k1k2)=n1NK=NK[/itex]

    But

    [itex] {\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=KNk_2=Kk_2N=KN[/itex]

    Where we used the facts that [itex]Nk_2=k_2N[/itex], [itex]n_1N=N[/itex] and [itex]Kk_2=K[/itex]. Therefore [itex]NK=KN[/itex] for all [itex]n\in N[/itex] and [itex]k\in K[/itex]
     
  5. Nov 15, 2012 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What does this string of equalities even mean? If we look at just the leftmost and rightmost expressions, you are asserting that
    [tex]n_1 k_1 n_2 k_2 = NK[/tex]
    The left hand side is an element, but the right hand side is a group. How can an element equal a group?
     
  6. Nov 15, 2012 #5
    Hmm, good point. I'll give this one last try using my method. Even if I don't get it this is good I feel like I'm learning a lot.

    Here is my modification that I hope fixes it.

    Let [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
    Then
    [itex]{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k_1k_2)=n_1nk=nk[/itex]

    But

    [itex] {\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=knk_2=kk_2n=kn[/itex]

    I guess the flaw now is that we can't assume that [itex]nk_2=k_2n[/itex]. Well I'll just use your method cause this probably won't work.
     
  7. Nov 15, 2012 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, this is better. But the problem is that the n and k in this equation:
    need not be the same as the n and k in this equation:
    You have the right idea, but you are making it more complicated than it needs to be. Try similar manipulations with the idea I suggested earlier.
     
  8. Nov 15, 2012 #7
    I guess since the intersection of n and k is {e} for both of my n's of N and k's of K then intuitively nk=e and nk=e :tongue2: which implies nkn^-1k^-1=e and I'm done. I just need to formalize that idea.
     
  9. Nov 15, 2012 #8

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I didn't follow your argument. You have [itex]n \in N[/itex] and [itex]k \in K[/itex]. Now what can you say about the element [itex]nkn^{-1}k^{-1}[/itex]? Hint: what if you add some parentheses: [itex](nkn^{-1})k^{-1}[/itex]?
     
  10. Nov 15, 2012 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Where have you used the fact that N and K are normal subgroups of G?
     
  11. Nov 15, 2012 #10
    OK, let me try to make it a little more precise.

    Since [itex]N[/itex] is normal in [itex]G[/itex], any element [itex]k\in K[/itex] which also happens to be in [itex]G[/itex] since [itex]K[/itex] is a subgroup (the fact that K is normal isn't important until the next step) of [itex]G[/itex] we get [itex]knk^{-1}=n \implies nk=kn[/itex] for all [itex]n \in N[/itex].

    Since [itex]K[/itex] is normal in [itex]G[/itex], any element [itex]n\in N[/itex] which also happens to be in [itex]G[/itex] since [itex]N[/itex] is a subgroup (the fact that N is normal is only important in the last step) of [itex]G[/itex] we get [itex]nkn^{-1}=k \implies kn=nk[/itex] for all [itex]k \in K[/itex].

    However, the intersection of N and K can only be {e} therefore [itex]nk=e=kn \implies nkn^{-1}k^{-1}=en^{-1}k^{-1}=ee=e[/itex].

    Furthermore [itex](nkn^{-1})k^{-1} \in K[/itex] and [itex]n(kn^{-1}k^{-1}) \in N[/itex]



    Am I done? It seems like there should be a more concise way to say it regardless.
     
  12. Nov 15, 2012 #11
    Wait wait I get it now....

    If [itex](nkn^{-1})k^{-1} \in K[/itex] and [itex]n(kn^{-1}k^{-1}) \in N[/itex] where we have used the fact that N ad K are normal in G then [itex](nkn^{-1})k^{-1}=e[/itex]

    and

    [itex](nkn^{-1})k^{-1}=e \implies nk=kn[/itex]


    Took me a while but I got it.
     
  13. Nov 15, 2012 #12

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Abstract Algebra HW
  1. Abstract algebra (Replies: 4)

Loading...