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Thread: 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 $\displaystyle G$ is abelian it means $\displaystyle H$ is a normal subgroup. Form the factor group $\displaystyle G/H$. The order of this group is $\displaystyle n$. Therefore, any element raised to $\displaystyle n$ must be the identity. Let $\displaystyle x\in G$. Then look at $\displaystyle xH \in G/H$. If we raise it to the $\displaystyle n$ i.e. add it $\displaystyle n$ times (because we are using additive notation as opposed to multiplicative) it means $\displaystyle (xH) + ... + (xH) = H \implies (nx)H = H \implies nx \in H$.
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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 $\displaystyle x \in G,$ then $\displaystyle \frac{x}{n} \in G,$ and hence $\displaystyle x=n \cdot \frac{x}{n} \in nG.$ so $\displaystyle G \subseteq nG.$ the fact that $\displaystyle nG \subseteq G$ is trivial. thus $\displaystyle nG=G.$

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