finding galois extension isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_4$ and $Q_8$ I am trying to find galois extension isomorphic to the groups listed above. I tried many things but they all land in the other 3 groups of order 8 that is $D_8,\mathbb{Z}_8,\ and \ \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.
 A: To construct a Galois extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}_4 \times \mathbb{Z}_2$, an idea would be to solve the easier problem of constructing a cyclic Galois extension $K/\mathbb{Q}$ of order $4$. Then simply adjoin some quadratic irrational $\sqrt{d} \not\in K$ to produce an order $8$ abelian extension $L = K(\sqrt{d})$. Since $L/\mathbb{Q}$ is generated by the Galois extensions $K/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{d})/\mathbb{Q}$, it is Galois. Furthermore, since $\textrm{Gal}(L/\mathbb{Q})$ has the order $4$ cyclic subgroup $\textrm{Gal}(K/\mathbb{Q})$ and two distinct order $2$ subgroups (the order $2$ subgroup of $\textrm{Gal}(K/\mathbb{Q})$ and the subgroup $\textrm{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})$, we must have
$$\textrm{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}_4 \times \mathbb{Z}_2.$$
To give you a hint, let $\alpha = \sqrt{2 + \sqrt{2}}$. You can show that $\mathbb{Q}(\alpha)/\mathbb{Q}$ is a cyclic Galois extension of order $4$ (the details of this can be found here). Then just choose $\sqrt{d}$ appropriately.
This post provides a detailed solution to the problem of finding a Galois extension $L/\mathbb{Q}$ with Galois group $Q_8$.
A: I think it is better to put the $C_8$ and $C_4 \times C_2$ cases apart (see below) and start from a biquadratic extension, i.e. a Galois  extension $K/\mathbf Q$ with group $G\equiv C_2 \times C_2$. Concretely, $K$ will be of the form $\mathbf Q(\sqrt a, \sqrt b)$, with $a$ and $b\in \mathbf Q^*$, and $ab^{-1}\notin \mathbf Q^{*2}$. Then construct a quadratic extension $L= K(\sqrt d)$, with $d\in K^*$ but $\notin K^{*2}$. A necessary and sufficient condition for $L/\mathbf Q$ to be normal is that $\sigma (d). d^{-1} \notin K^{*2}$ for any $\sigma \in G$ (easy Galois exercise). Actually, it is enough to check this property for two generators of $G$. 
Once the Galois extension $L/\mathbf Q$ of degree 8 is constructed, you have to decide whether its Galois group $E$ is $C_2^{3}$, or $D_8$, or $Q_8$. Since you have discarded $C_8$ and $C_4 \times C_2$, the only abelian case is $L =\mathbf Q(\sqrt a, \sqrt b, \sqrt c)$, with $c\in \mathbf Q^*$ and $a, b, c$ independent mod $\mathbf Q^{*2}$. To distinguish between the two non abelian cases $D_8$ and $Q_8$, just note that they are characterized by the number of their cyclic subgroups of order 4: $D_8$ admits exactly 1 such subgroup; as for $Q_8$ , all its 3 subgroups of order 4 are cyclic. So introduce the 3 quadratic subfields $k_i$ of $K$ and  compute the 3 Galois subgroups $Gal(L/k_i)$ (a tedious task). If lucky, you don't  need to do the complete calculations, because there is a criterion which says that $L/k_i$ is embeddable into a cyclic extension of degree 4 iff −1 is a norm in $L/k_i$ (see below).
To justify the above assertion, you can appeal to a general proposition: given a field $k$ of characteristic $\neq 2$, containg a primitive $2^n$-th root $\zeta$ of $1$, a cyclic extension $K/k$ of degree $2^n$ is embeddable into a cyclic extension $L/k$ of degree $2^{n+1}$  iff $\zeta$ is a norm in $K/k$. This can even be generalized, replacing 2 by any prime $p$. See e.g. https://math.stackexchange.com/a/1691332/300700. Finally,the $C_8$ case is akin: construct $L/\mathbf Q$ cyclic of degree 8 by successive quadratic extensions and using the previous criterion. Analogously for $C_4$ (this completes the answer of @ Ethan Alweise), from which $C_4 \times C_2$ follows immediately.
