Let $r>0$. I want to prove that there exists a extension $K / \mathbb Q_2$ such that $Gal(K/\mathbb Q_2)= \mathbb Z/2 \times (\mathbb Z /2^r)^2$. This is an exercise in Kedlaya's notes on class field theory see here.
Here is what I thought. Let me know if it looks good:
By local class field theory $$ \widehat {\mathbb Z} \times \mathbb Z_2 ^* \simeq Gal (\mathbb Q_2 ^{ab}/\mathbb Q_2)$$ Since $\mathbb Z_2^*$ has a subgroup with quotient $\mathbb Z/2 \times \mathbb Z/2^r$ and $\widehat {\mathbb Z}$ has a subgroup with quotient $\mathbb Z/2^r$. By taking product of these subgroup, the corresponding field $K$ is an abelian extension of $\mathbb Q_2$ with the required Galois group.
Furthermore, we can find this $K$ explicitly. The field corresponding to the quotient $\mathbb Z/2 \times \mathbb Z/2^r$ is obtained by adjoining $\zeta _{2^{r+2}}$ (so this is the totally ramified extension) and the field corresponding to the quotient $\mathbb Z/2^r$ is obtained by adjoining $\zeta _{2^{2^r}-1}$ (this is the unramified extension). So $K= \mathbb Q_2(\zeta _{2^{r+2}}, \zeta _{2^{2^r}-1})$. Is this correct? The hint given in the notes appears to be incorrect.
Your solution is correct and indeed the hint is incorrect. In general, degree $f$ unramified extensions of $\mathbb Q_p$ is obtained by adjoining a primitive $(p^f-1)$-th root of unity, as can be checked from Hensel's lemma. Thus, the field $L$ that Kedlaya suggests contains $\mathbb Q_2(\zeta_{2^r-1})$, a degree $r$ unramified extension over $\mathbb Q_2$. Of course, $r$ need not divide $[L:\mathbb Q_2]=2^{2r+1}$ so this cannot be correct.