# The group $\mathbb{Q}^\times / \left( \mathbb{Q}^\times \right)^2$

Does there exist a nice description of the group $\mathbb{Q}^\times / \left( \mathbb{Q}^\times \right)^2$? What is its order?

Here, we see that $\mathbb Q^{\times}\cong \mathbb Z/2\mathbb Z\oplus\left(\bigoplus_{i=1}^{\infty}\mathbb Z\right)$. Continuing to write additively, $$2\left(\mathbb Z/2\mathbb Z\oplus\left(\bigoplus_{i=1}^{\infty}\mathbb Z\right)\right)=\bigoplus_{i=1}^\infty\mathbb 2\mathbb Z.$$ More precisely, the multiplication by $2$ turns a tuple $(a,b,c,\dots)$ with almost all terms zero into the tuple $(0,2b,2c,\dots)$.
It follows that $$\mathbb Q^\times/2\mathbb Q^\times\cong\mathbb Z/2\mathbb Z\oplus\bigoplus_{i=1}^\infty\mathbb Z/2\mathbb Z\cong\bigoplus_{i=1}^\infty\mathbb Z/2\mathbb Z.$$
• I don't understand... you have found the isomorphism-type of $\mathbb{Q}^\times / 2\mathbb{Q}^\times$, not of $\mathbb{Q}^\times / \left( \mathbb{Q}^\times \right)^2$. – Frankenstein Jul 31 '13 at 0:56
• @Frankenstein: How so? $2{\bf Q}^\times$ is just the same as ${\bf Q}^\times$, so the quotient you speak of is trivial. – tomasz Jul 31 '13 at 1:12
• @Frankenstein I am writing everything additively in my post since the actual computations involve abelian groups that are usually written additively. Hence, my $\mathbb Q^\times/2\mathbb Q^\times$ is the same as your $\mathbb Q^\times/(\mathbb Q^\times)^2$. I will second tomasz's point that I did not find the iso-type of $\mathbb Q^\times/2\mathbb Q^\times$ as you would write it. – user714630 Jul 31 '13 at 1:22