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Basic Group Theory Question

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Let (R^2,+) be the set of ordered pairs with addition defined component wise. Verify {(x,2x)|x£R} is a subgroup and that {(x,2x+1)|x£R} is not a subgroup.



    3. The attempt at a solution
    So for something to be a subgroup it has to have all it's set items contained in the main group and be a group on it's own.

    To be a group on it's own it has to have an identity e and an operation (in this case +) so that:
    e+g = G

    and an inverse so that (in this case a negative number with same magnitude)
    g + (-g) = 0

    and be assosciative
    (g+h)+k = g+(h+k)
    and it seems both (x,2x) and (x,2x+1) satisfy all these properties.

    Furthermore, it seems that both (x,2x) and (x,2x+1) have all their combinations included in the main group. How is (x,2x+1) not a subgroup? What am I missing here?
     
  2. jcsd
  3. Jun 6, 2014 #2

    HallsofIvy

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    I presume you mean e+ g= g.

    Since you are using ordinary addition, (x, 2x+ 1)+ (y, 2y+ 1)= (x+ y, 2x+1+ 2y+ 1)= (x+ y, 2(x+ y)+ 2) which is NOT of the form (a, 2a+ 1).
     
  4. Jun 6, 2014 #3

    AlephZero

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    You are missing the fact that if H is a subgroup of G, and x and y are any elements of H, then x+y must also be an element of H.
     
  5. Jun 6, 2014 #4

    pasmith

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    The operation of the subgroup is the same as the operation of the full group. Thus you already know that the operation is associative, that there is a unique identity [itex]e \in G[/itex], and that every [itex]g \in G[/itex] has a unique inverse [itex]g^{-1} \in G[/itex].

    To show that [itex]H \subset G[/itex] is a subgroup, you must show the following (in order of being easiest to check):
    • Identity: [itex]e \in H[/itex].
    • Inverse: If [itex]h \in H[/itex] then [itex]h^{-1} \in H[/itex].
    • Closure: If [itex]h_1 \in H[/itex] and [itex]h_2 \in H[/itex] then [itex]h_1h_2 \in H[/itex].

    Here [itex]H = \{(x, 2x + 1) : x \in \mathbb{R}\}[/itex] falls at the first hurdle: [itex](0,0) \notin H[/itex].
     
  6. Jun 6, 2014 #5

    Zondrina

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    Pasmith has noted something important in his post. If the origin isn't included, then it can't be a subgroup. Any line with a y intercept that isn't 0 can't form a subgroup.

    So ##(x,2x)## will form a subgroup, but ##(x,2x + b)## for any ##b ≠ 0## will not.
     
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