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Math Help - Cosets

  1. #1
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    Cosets

    The questions are
    Let G be the group of rational numbers under addition and n a positive integer.
    a) Suppose H is a subgroup of G with index n. Show that nx is an element of H for all x in G.

    I think this has something to do with nx=x+x...+x (n times) and somehow it relates to G and the fact that there is only one coset of H in G.

    b) Let nG={nx element G : x element G} Show that nG=G. Hence or otherwise, prove that G has no proper subgroup of finite index.

    I think if I set F to be the group of reals under addition then G is a subgroup of F. So nG can be viewed as a coset of G in F and since n is in G then nG=G (that being a property of a coset) Is that right?
    But how do I show that G has no proper subgroup of finite index?
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  2. #2
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    Quote Originally Posted by Rivan View Post
    a) Suppose H is a subgroup of G with index n. Show that nx is an element of H for all x in G.
    Since G is abelian it means H is a normal subgroup. Form the factor group G/H. The order of this group is n. Therefore, any element raised to n must be the identity. Let x\in G. Then look at xH \in G/H. If we raise it to the n i.e. add it n times (because we are using additive notation as opposed to multiplicative) it means (xH) + ... + (xH) = H \implies (nx)H = H \implies nx \in H.
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  3. #3
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    Quote Originally Posted by Rivan View Post
    b) Let nG={nx element G : x element G} Show that nG=G.
    if x \in G, then \frac{x}{n} \in G, and hence x=n \cdot \frac{x}{n} \in nG. so G \subseteq nG. the fact that nG \subseteq G is trivial. thus nG=G.

    Hence or otherwise, prove that G has no proper subgroup of finite index.
    suppose H \subseteq G is a subgroup of finite index n. then by part a): nG \subseteq H and by the first part of b): nG=G. thus G \subseteq H, i.e. H=G. \ \ \ \square
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