Existence of Galois group of order 8 as $\mathbb Z_2\times \mathbb Z_4$ I'm trying to find a Galois group of $\mathbb Z_2\times \mathbb Z_4$ but, first, I'd like to know if it does exist. I know that a Galois group of order $8$ must be a subgroup of the symmetric group of order $4$, $S_4$. Also, this subgroup must be transitive.
Hence, I should see that $\mathbb Z_2\times \mathbb Z_4$ is not transitive. But how could I do that?
 A: The simplest example is probably the polynomial $X^8+1$, and now I’ll explain why.
It’s a fact that every finite abelian extension of $\mathbb Q$ is contained in a cyclotomic extension $\mathbb Q(\zeta_m)$, $\zeta_m$ being a primitive $m$-th root of unity. The Galois group of $\mathbb Q(\zeta_m)$ over $\mathbb Q$ is the group of units of the ring $\mathbb Z/m\mathbb Z$. With this information, you can get any finite abelian group as the quotient of a $(\mathbb Z/m\mathbb Z)^\times$, and so an extension of $\mathbb Q$ with that Galois group.
Now, $\mathbb Z/16\mathbb Z$ has for its units the odd numbers modulo $16$, as a multiplicative group, and you easily check that this has the shape $\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z$, generators being $-1$ and $5$. The polynomial for the primitive sixteenth roots of unity is $(X^{16}-1)/(X^8-1)=X^8+1$, and there you have it.
A: Consider the splitting field of $x^5-1$, $x^2-2$ on $\mathbb{Q}$.
A: A) The Galois group of a degree $n$ polynomial need not be a transitive subgroup of $S_n$. E.g. what is the Galois group of a biquadratic extension? What you have in mind is that the Galois group of an irreducible polynomial is a transitive subgroup of $S_n$, where $n$ is the degree of the polynomial.
B) Why does a Galois group of order 8 have to be a subgroup of $S_4$? What tells you that you cannot have, say, a cyclic group of order 8 as a Galois group (which would then be a transitive subgroup of $S_8$)?
So with these remarks in mind, perhaps you should ask the precise question that you meant to ask.
Anticipating what your question might be: MAGMA tells me that the group generated by
$$
(1, 4)(2, 5)(3, 8)(6, 7),\\ (1, 5, 6, 8)(2, 7, 3, 4),\\ \text{and } (1, 6)(2, 3)(4, 7)(5, 8)
$$ is a transitive subgroup of $S_8$ that is isomorphic to $C_2\times C_4$. It also tells me that there are no such subgroups in any $S_n$ for $n$ smaller than 8.
A: Actually, the simplest example is $\Bbb Q(\zeta_{15})$, since cyclotomic extensions are well-known.
A quick calculation shows $\operatorname{Gal}(\Bbb Q(\zeta_{15})/\Bbb Q) = (\Bbb Z/15\Bbb Z)^\times = (\Bbb Z/3\Bbb Z)^\times \times (\Bbb Z/5\Bbb Z)^\times = C_2 \times C_4$.
