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Abstract Algebra. Need help!

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Let a be in a group G, and let

    [tex] H=\{ a^n: n\in Z\}[/tex]. Show the following:

    (i) if h and h' are in H, so is hh'.

    (ii) The identity e of G is in H.

    (iii) if h is in H, so is [tex] h^{-1}[/tex].

    3. The attempt at a solution

    Here is what i tried. First of all i am not sure that what i have done is correct, second i am not sure how a bein an element of G helps here. However i have tried to make use of such a fact in one or another way. Let's see:


    (i) Let h and h' be in H, then both of them are of the form : [tex] h=a^{n_1} ; \ \ \ h'=a^{n_2}[/tex] where [tex]n_1,n_2\in Z[/tex] .

    Now, since: [tex] a\in G=> a^{n_1},a^{n_2}\in G[/tex]

    Then :

    [tex] hh'=a^{n_1}a^{a_2}=a^{n_1+n_2}\in H[/tex], since [tex] n_1+n_2\in Z[/tex]. ( Note: This is also where i am not sure wheter i am properly making use of the fact that a is in G, because if a wouldn't be in G, then we would not be sure that we can have [tex] a^{n_1} [/tex] or [tex] a^{n_2}[/tex] right? or?


    (ii) I am not sure on this one at all, here is what i said:

    let [tex] e=a^0[/tex] be the identity element in G, so [tex] e=a^0\in H[/tex] since [tex] n=0\in Z[/tex]

    (iii) Since [tex]h\in H[/tex] then h is of the form [tex] h=a^n\in H[/tex] for some n in Z. But also [tex] a^{-n}\in H[/tex] since [tex]-n\in Z[/tex]. So we have:

    [tex] a^{-n}=(a^n)^{-1}=h^{-1}\in H[/tex], again here i think , that if a would not be an element of the group G, then we wold not be able to perform the following step, that i did on my 'proof' : [tex] a^{-n}=(a^n)^{-1}[/tex] right?




    I would appreciate any further and more detailed explanation from you guys. Even if i am pretty close, please try to give me further explanations.

    Thanks in advance!
     
  2. jcsd
  3. Sep 8, 2008 #2
    Here's what I would say:

    I don't have my abstract algebra book handy, but if I remember the properties of groups well enough:

    i ) looks okay to me.

    ii) I would say something like: Since G is a group, if a is in G then a[tex]^{-1}[/tex] is in G. Now a = a[tex]^{1}[/tex] [tex]\in[/tex] H and a[tex]^{-1}[/tex] [tex]\in H[/tex] since 1 and -1 are integers. But a*a[tex]^{-1}[/tex] = a[tex]^{0}[/tex]= e, thus e is in H.

    iii) Similarly:

    Let h be in H. Then h = a[tex]^{n}[/tex] for some n in Z. Now consider h' = a[tex]^{-n}[/tex], which is in H since -n is in Z.
    Now h*h' = a[tex]^{n-n}[/tex] = a[tex]^{0}[/tex] = e. (It's trivially true the other way, h'*h=e). Since h*h' = h'*h = e, h' is the inverse of h.
     
  4. Sep 9, 2008 #3
    Well, for part (ii) of this problem, here is what my prof said.

    Since he said we have to assume that G, is finite, then it means that the elements of G will start to repeat after some time. That is

    [tex] a^m=a^n[/tex] for some m,n in Z. then these two el. are also in H. Also there exists [tex] (a^n)^{-1}[/tex] so multiplying from the right there we will get:

    [tex] a^{m-n}=e\in G[/tex] and also in Z.
     
  5. Sep 9, 2008 #4

    Dick

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    H is a subgroup even if you don't assume G is finite. a^0 is defined as e. And a^(-n) is defined as the inverse of a^n. If you meant n is an element of N (the POSITIVE integers), then you need an additional assumption, and yes, G finite will do it.
     
  6. Sep 9, 2008 #5
    Well, i guess we needed that extra assumption about G being finite, since we haven't learned subgroups yet, so i guess we couldnt make use of that fact, that H is a subgroup of G.
     
  7. Sep 9, 2008 #6

    Dick

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    I guess the real question is what did you mean by Z? Is it all integers, or only the positive ones?
     
  8. Sep 9, 2008 #7
    Well, in the book they did not specify it, but i guess they meant all integers.
     
  9. Sep 9, 2008 #8

    Dick

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    Then you don't need that G is finite. I probably shouldn't have said "subgroup". But the properties you are trying to show imply H is a subgroup of G.
     
  10. Sep 15, 2008 #9
    How would one prove that if h is in H, then also h^-1, that is its inverse of it is also in H, if Z is only positive integers.

    I managed to show, as stated that [tex] e=a^{m-n}\in H[/tex] since i supposed that the group G is finite, and i also know that now i have to take two elements in H, say h and h' and then use the property that H is closed, and

    [tex]hh'=...=e\in H[/tex], but i cannot figure out how to pick up h and h'?

    Can u help me on this?
     
  11. Sep 16, 2008 #10
    Re: Abstract Algebra. I STILL Need help on this one!

    wwwww
     
  12. Sep 16, 2008 #11
    I think i got it, let's see:

    SInce [tex] g^{m-n}=e\in H[/tex] it means that H consists of m-n elements, that is all powers of g up to m-n. Now this means that there defenitely are:

    [tex] s_1, s_2 \in Z^{+}[/tex] such that [tex] s_1+s_2 =m-n[/tex] while [/tex] g^{s_1},g^{s_2} \in H[/tex] this means that:

    For every element in H, say [tex] h=g^{s_2}, s_2 \in Z^+, \exists h'=g^{s_1}, s_1 \in Z^+[/tex] such that

    using closure property of H, we get:


    [tex] hh'=g^{s_1}g^{s_2}=g^{s_1+s_2}=g^{m-n}=e \in H[/tex] which means that h and h' are inverses of each other, and still both h and h' are in H. Proof done. RIght?
     
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