Let $G$ be a finite abelian group. Show that there is a Galois extension $K/\Bbb Q$ with $\text{Gal}(K/\Bbb Q) \cong G$.
I have seen a proof using that for a fixed positive integer $n$, there are infinitely many prime numbers $p$ such that $p ≡ 1 \bmod n$.
But is there any easy way to prove this? Please reply
Yes there is a simple argument to that special case of Dirichlet theorem, and it is a corollary of the following elementary fact:
Given $f(x) \in \mathbb{Z}[x]$ non constant, than the set of primes $q$ such that $f(x)$ has a linear factor in $\mathbb{F}_q[x]$ is infinite.
You can prove this along the same kind of argument of Euclid: list those elements, say $p_1,...,p_n$, then observe that the constant factor, that is, $f(0)$ is factorized just by these primes, so deduce that the expression $f((p_1\cdots p_n)^k)$ has the same $p_i$-valuation of $f(0)$ for $k$ large enough, so you get that $f((p_1...p_n)^k)$ is bounded with $k$ varying, so $f$ is constant, contradiction.
Now apply this to $\Phi_n(x)$, the $n$-th cyclotomic polynomial, plus the observation that if $x-a\mid \Phi_n(x)$ in $\mathbb{F}_q[x]$ then $\mathrm{ord}_q(a)=n$ so $n\mid q-1$.
