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Subset of the integers

  1. Aug 2, 2007 #1
    Hello all.

    I reluctantly ask this question because it is probably,as the text states easy, but my desire to clear this point up overides my fear of looking a fool.

    I quote word for word but will use words instead of the belongs to symbol.

    A subset S of the set Z of integers is a subgroup of Z if 0 is in S, -x is in S,and x+y is in S for all x and y in S. It is easy to see that a non empty subset S of Z is a subgroup of Z if and only if x-y is in S for all x and y in S.

    I understand the definition and I can see that 0 is in S. I can only assume that somehow because -x ( the additive inverse of x ) is in S that this guarantees that x is in S. If S were a subgroup of Z of course -x being in the subgroup means that x was in it. But we are not assuming that S is a subgroup but testing for it.

    Help!

    Matheinste.
     
  2. jcsd
  3. Aug 2, 2007 #2

    radou

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    Well, first of all, how "can" you see that 0 is in S? S is non empty. So a is in S. And you know that aa^-1 = e ( = 0) is in S.
     
  4. Aug 2, 2007 #3
    Thaks radou. I am not being deliberately awkward, I just need to completely grasp the basics which I used to find uninteresting but now regard as most interesting. As S is defined only as a subset ( non empty } and has therefore no structure defined I cannot see how we can assume anything about the set from its being non empty.

    More help please.

    Matheinste.
     
  5. Aug 3, 2007 #4

    radou

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    Ok, what exactly is your question? Does this bother you?

    If so, you already know that 0 is in S, from the previous post. Now, (since we're proving direction "<=", and we assumed that x-y is in S for all x and y in S), what can we "do" with 0 and a?
     
  6. Aug 3, 2007 #5
    Thanks radou. I think I may be a little less stupid today.

    I see now that if we have at least one element in S and call it x then we have x-x=0 is in S by what we are given. Then we can have 0-x is in S and so
    -x is in S and this is our inverse element and so x-(-x)=x+x is in S. This fulfils the requirements stated subgroup.

    Is that OK.

    Matheinste.
     
  7. Aug 3, 2007 #6

    HallsofIvy

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    I'm puzzled by that last statement. Why is it important that x+ x be in S?
    More to the point is that if x and y are in S, then so is -y and x-(-y)= x+ y is in S.
     
  8. Aug 3, 2007 #7
    Thankyou HallsofIvy. I take note of what you have said. I am now happy with the answer.

    Matheinste.
     
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