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Algebraic structures (a subgroups proof)

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Let n>1 be an integer, and let a be a fixed integer, prove or disprove that the set
    E will stand for is an element of

    H = {xEZ| ax = (mod n) is a subgroup under addition.

    3. The attempt at a solution

    I recognized that ax = a + a + a... + a. I figured this might help me along the way,s ince I am trying to prove ax a subgroup under addition.

    Assume H is a subgroup under addition.

    (From here I hope to prove that H fulfills all of the prerequisites of a subgroup (1. non-empty 2. is closed 3. every number has an inverse). If it does not, by contradiction I can prove that H is not a subgroup under addition.)

    1. Since a is a fixed integer and is contained contained in the set H, we know that H is non-empty.

    I'm not too sure what to do from here.

    2. Since ax = 0(mod n), ax = nq for some q.

    Do we need to prove that a and x are in H at all, since the set cannot be closed if they aren't... it only says ax is contained in H.

    then we would need to show that a + a + a ... + a = 0(mod n) is contained in H.

    Could we possibly use inductions to solve this?

    thanks for the help in advance
  2. jcsd
  3. Dec 6, 2008 #2
    There's something missing here. Do you mean [tex]H = \{ x \in \mathbb{Z} \mid ax \equiv 0 \mod n \}[/tex]?

    Where does it say this? If n = 2 and a = 1, then H would be the set of all even integers and a isn't in H.
  4. Dec 6, 2008 #3
    thank you for adding that } that I forgot... I don't exactly know how to use latex, so I'm not sure how to express congruence. Either way, it is clear you understand the problem.

    I'm not really sure at all what that second statement is saying... perhaps you could elaborate a little.
  5. Dec 6, 2008 #4
    Make sure you understand what H contains. The given definition of H says that H contains all integers x where ax is congruent to 0 mod n. It's not said that H contains all integers ax congruent to 0 mod n. It's not said that H contains a. These last two claims may be true or false; you have to figure them out.

    For the three conditions,

    1. Show that some element is in H. Since this element has to be in H regardless of what n and a are, it has to be pretty versatile. Think of the most common element that is in every group. Think of the element in every subgroup of [tex]\mathbb{Z}[/tex].

    2. If x and y are in H, what conditions do they satisfy? What does this tell you about x + y?

    3. Same question for -x.
  6. Dec 7, 2008 #5
    1. Well, if i said that the element was the empty set, I suppose i would correct ikn saying that it is in every group; however, isn't the set of the empty set considered not considered non-empty?
    I can't think of another element in every subgroup of Z.
    ooo, you mean the identity element.

    2. I guess if x and y are in H, then they both must be divisible by n.

    so x = qn and y = pn for some q,p E Z.
    -> x + y = pn + qn
    -> (x + y) = n(p + q)
    -> n|(x+y)
    Last edited: Dec 7, 2008
  7. Dec 7, 2008 #6
    1. Right. The empty set is a subset, not an element, of subsets of Z.

    2. No, x and y don't have to be divisible by n. It's not x = 0, but ax = 0 mod n.
  8. Dec 7, 2008 #7
    so do i go from ax = nq for some q E Z....

    i honestly have no idea how to start this.

    I freeze when I get these proofs. And I need to do something like 4 of them.
  9. Dec 7, 2008 #8
    You have the right idea, and writing ax = nq etc. should certainly work.

    With ax = 0 and ay = 0, there is an easier way to show that a(x + y) = 0.
  10. Dec 7, 2008 #9
    yea... i cant get them to equal 0.
    I have a feeling that I'm completely off track.

    I tried ax = 0(mod n) implies ax = nq for some q E Z.
    So we also know that ax - nq = 0.
    suppose ax = 0 and ay = 0 for some x,y E Z.
    ax = ay
    ax - ay = 0
    ax - ay = ax - nq
    ax + nq = ax + ay
    nq + nq = ax + ay
    2nq = a(x + y)

    I'm not really sure how proving that 0 = (x + y) brings us any closer to proving that H is closed either.
  11. Dec 7, 2008 #10
    That is the right idea.

    Remember the goal is to show that if [tex]x, y \in H[/tex], then [tex]x + y \in H[/tex]. Once you show that a(x + y) [tex]\equiv[/tex] 0 (mod n), then you have the desired conclusion.

    When you write this down, be careful with the difference between = and [tex]\equiv[/tex] (mod n). You used "=" for both senses above, so it can be confusing.
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