Embedding of a field in a cyclic extension Show that  

$K=\mathbb{Q}(\sqrt {a})$ for $a\in \mathbb{Z}$, $a<0$ can not  be embedded in a cyclic extension whose  degree over $\mathbb{Q}$ divisible by 4.

I have tried for order exactly $4$ and got the answer but that is not so conceptual... I felt i am not using full power of cyclic group property...
There is another post Embedding of a field extension to another with same question but i could not get the idea that has been said in that question... I can not ask the person who has answered this as he is no more an active member and i can not ask this OP to edit this question as he/she is also not active...
So, please do not close this as duplicate...
So, please help me to understand this better...
 A: The fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with Galois group $G=\mathrm{Gal}(L/F)$, there is an order-reversing, degree-preserving bijection
$$\{\text{intermediate fields }L\supseteq E\supseteq F\}\longleftrightarrow\{\text{subgroups }G\supseteq H\supseteq\{e\}\}$$
where the map from left to right is 
$$E\longmapsto \mathrm{Fix}(E)=\{\sigma\in G:\sigma|_E=\mathrm{id}_E\}$$
If $L/\mathbb{Q}$ is a cyclic extension of degree divisible by $4$, this means $L/\mathbb{Q}$ is a Galois extension with
$$\mathrm{Gal}(L/\mathbb{Q})\cong C_n,\qquad 4\mid n$$
One basic property of $C_n$ is that for any $d\mid n$, the number of elements of order $d$ is precisely $\varphi(d)$, where $\varphi$ is Euler's totient function. In particular there is only one element of order $2$ in $\mathrm{Gal}(L/\mathbb{Q})$.
Because $a<0$ the number $\sqrt{a}$ is imaginary, and $\sqrt{a}\in L$ so that complex conjugation $\rho:L\to L$ is among the elements of $\mathrm{Gal}(L/\mathbb{Q})$. Obviously $\rho$ is an element of order $2$, so it must be the unique such element. However, the intermediate field
$$L\supseteq\mathbb{Q}(\sqrt{a})\supseteq\mathbb{Q}$$
has a corresponding subgroup $\mathrm{Gal}(L/\mathbb{Q})\supseteq H\supseteq\{0\}$ with
$$|H|=[L:\mathbb{Q}(\sqrt{a})]=\frac{n}{2}$$
Because $n$ is divisible by $4$, this is still even, and $H$ is a cyclic group (because it is a subgroup of a cyclic group) so that $H$ contains an element of order $2$. Thus it must contain $\rho$, the only element of order $2$ in $\mathrm{Gal}(L/\mathbb{Q})$. But
$$H=\{\sigma\in \mathrm{Gal}(L/\mathbb{Q}):\sigma|_{\mathbb{Q}(\sqrt{a})}=\mathrm{id}_{\mathbb{Q}(\sqrt{a})}\}$$
cannot contain $\rho$ because complex conjugation does not fix $\sqrt{a}$; contradiction.
A: Suppose you find an embedding $\Bbb Q \subset K \subset M$ where $G = Gal(M/\Bbb Q)$ is cyclic of order $4n$.
By the fundamental theorem of Galois theory, intermediate fields between $\Bbb Q$ and $M$ correspond to subgroups of $G$.
As a cyclic group of order $4n$, $G$ has exactly one subgroup $H_2$ of index $2$ ,so $M$ has exactly one quadratic subfield, which has to be $K$, and exactly one subgroup $H_4 \subset H_2 \subset G$ of index $4$, corresponding to a unique subfield $\Bbb Q \subset K \subset L \subset M$ where $Gal(L/\Bbb Q) = G/H_4 = \Bbb Z/4 \Bbb Z$.
If you have proved you can't embed $K$ into a cyclic extension $L$ of degree $4$ of $\Bbb Q$, it follows that you can't embed $K$ into a cyclic extension of degree $4n$.
A: A quadratic field K can be embedded in a cyclic extension of degree 4 iff -1 is a norm in K/Q. See the question Conceptual reason why a quadratic field has $-1$ as a norm if and only if it is a subfield of a $\mathbb{Z}/4$ extension? first posed by Ben Blum-Smith on Dec. 6'13 –  Here K is imaginary quadratic, hence every norm is positive, and -1 cannot be a norm.
