What are the elements of $(\mathbb{Q}_{2}/(\mathbb{Q}_{2})^{2})^{\times}$? The answer is $\{-1, \pm 2,\pm 5, \pm 10\}$ and I can't even figure out why 10 is there.
I mean the 2-adic unit rationals. It turns out that $\mathbb{Q}_{2}^{\times}/(\mathbb{Q}_{2}^{\times})^{2}\simeq (\mathbb{Z}/2\mathbb{Z})^{3}$. Let's start with the elements of $(\mathbb{Q}_{2})^{\times}$.  What are they? Is $\pm 10$ one of them?
 A: Every nonzero element of $\Bbb Q_p$ may be written as $p^nu$ for some $n\in\Bbb Z$ and $u\in\Bbb Z_p^\times$. Use this fact to prove that $\Bbb Q_p^\times\cong \Bbb Z\times \Bbb Z_p^\times$. Can you tell what the isomorphism is? Note this holds for any $p$.
The subgroup of squares corresponds to $2\Bbb Z\times(\Bbb Z_p^\times)^2$. (How?) You can use Hensel's lemma to prove that $(\Bbb Z_2^\times)^2$ is precisely $1+2^3\Bbb Z_2$ (note this works out differently for odd primes). This is precisely the kernel of the map $\Bbb Z_2^\times\to (\Bbb Z/2^3\Bbb Z)^\times$ (the mod $8$ map), so to get a system of representatives we merely need to pick representatives for elements in $(\Bbb Z/2^3\Bbb Z)^\times$, the most obvious is $\{1,3,5,7\}$ but this can also be written $\{\pm1,\pm5\}$. For the $\Bbb Z$ component of $\Bbb Q_2^\times$, the powers of $2$, the system of representatives are $0$ and $1$ which correspond to $2^0$ and $2^1$ multiplicatively in $\Bbb Q^\times_2$. Therefore,
$$\frac{\Bbb Q_2^\times}{(\Bbb Q_2^\times)^2}\cong\frac{\Bbb Z}{2\Bbb Z}\times\left(\frac{\Bbb Z}{2^3\Bbb Z}\right)^\times\cong\left(\frac{\Bbb Z}{2\Bbb Z}\right)^3$$
with a system of representatives given by $\{2^0,2^1\}\cdot\{\pm1,\pm5\}=\{\pm1,\pm2,\pm5,\pm10\}$.
