1. ## 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?

2. Originally Posted by Rivan
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 $(xH) + ... + (xH) = H \implies (nx)H = H \implies nx \in H$.

3. Originally Posted by Rivan
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. $H=G. \ \ \ \square$