November 13th
Today I learned that the inverse Galois problem holds for abelian groups, and it's actually fairly nice. Fix a finite abelian group $A.$ The restriction that we make is to try to find $A$ as an extension over $\QQ,$ and the rest of the proof is forced.
From Kronecker-Weber, we know that every abelian extension is a subextension of a cyclotomic extension, so we look inside cyclotomic extensions. These all have Galois groups $(\ZZ/N\ZZ)^\times,$ so if our approach is possible, we had better be able to fit $A$ inside of such a group. However, the only way to access $(\ZZ/N\ZZ)^\times$ is via Chinese Remainder Theorem. With this in mind, we would like to express $A$ as a product of cyclic groups, so we invoke the classification of finite abelian groups to write\[A\cong\prod_{k=1}^N\ZZ/n_k\ZZ\]for some sequence of integers $\{n_k\}.$ For example, the classification of finite abelian groups gives $n_k$ each a power of a prime.
Now we proceed with the construction. We can actually show (classically) that there are infinitely many primes $1\pmod{n_k}$ for any of the $k,$ so we may fix $\{q_k\}$ a sequence of strictly increasing primes such that $q_k\equiv1\pmod{n_k}.$ This technically follows from Dirichlet's theorem, so we don't talk about this more here. Let $Q$ be the product of these primes, and now we see\[(\ZZ/Q\ZZ)^\times\cong\prod_{k=1}^N(\ZZ/q_k\ZZ)^\times\cong\prod_{k=1}^N\ZZ/(q_k-1)\ZZ.\]In particular, we $n_k\mid q_k-1$ implies that\[A\cong\prod_{k=1}^N\ZZ/n_k\ZZ\cong\prod_{k=1}^n\frac{\ZZ/(q_k-1)\ZZ}{n_k\ZZ/(q_k-1)\ZZ}\cong(\ZZ/Q\ZZ)^\times\bigg/\prod_{k=1}^Nn_k\ZZ/(q_k-1)/\ZZ.\]The last isomorphism holds by looking at the kernel of $\ZZ/Q\ZZ$ onto the middle group. So we let $B$ be the subgroup giving $A\cong(\ZZ/Q\ZZ)^\times/B.$ Therefore we have fit $A$ inside of $(\ZZ/Q\ZZ)^\times.$
The rest of the finish is Galois theory. We know $\op{Gal}(\QQ(\zeta_Q)/\QQ)\cong(\ZZ/Q\ZZ)^\times,$ and from the Galois correspondence, every subgroup of $(\ZZ/Q\ZZ)^\times$ appears as a fixed field. Let $K$ be the fixed field corresponding to $B.$ Then, because our extension is abelian, every subgroup and therefore every extension is normal, so we see\[\op{Gal}(K/\QQ)\cong\op{Gal}(\QQ(\zeta_Q)/\QQ)/B\cong(\ZZ/Q\ZZ)^\times/B\cong A.\]This is what we wanted, so we're done here.