Galois group of a biquadratic quartic From Hungerford, section V, chapter 4 exercise 9:

Let $x^4+ax^2+b$ in  $K[x]$ (with char $K\neq $2) be irreducible with
  Galois group $G$.
(a) If $b$ is a square in $K$, then $G =
 \mathbb{Z}_2\times\mathbb{Z}_2$.
      (b) If $b$ is not a square in $K$
  and $b(a^2-4b)$ is a square in $K$, then $G = \mathbb{Z}_4$.
      (c)
  If neither $b$ nor $b(a^2-4b)$ is a square in $K$, then $G = D_4$
  (dihedral group).

What I've tried
With (a):
 Let be $u_1,u_2,u_3,u_4$ the roots of the quartic, then $b=u_1u_2u_3u_4$ is in $K$ and there exists $a\in K$ with $a^2=b$. On the other hand, $G = \mathbb{Z}_2\times\mathbb{Z}_2$ if and only if $\alpha=u_1u_2+u_3u_4$, $\beta=u_1u_3+u_2u_4$, $\gamma=u_1u_4+u_2u_3$ are in $K$, but I don't find the way to rely those results. 
With (b) I know that $b=u_1u_2+u_1u_3+u_1u_4+u_2u_3+u_2u_4+u_3u_4$ but again I dont know how to continue.
Any help? Thanks. 
 A: There are two numbers $\alpha,\beta$ (in some algebraic closure of $K$)
such that the identity $P=X^4+aX^2+b=(X^2-\alpha^2)(X^2-\beta^2)$ holds. Then the 
set $R$ of all roots of $P$ is $\lbrace \pm \alpha, \pm \beta\rbrace$.
Since $P$ is irreducible, $G$ can be identified to a transitive subgroup
of ${\mathfrak S}(R)$, the group of permutations of $R$. 
Also, any $\sigma\in G$ obviously satisfies $\sigma(-\alpha)=-\sigma(\alpha)$
and $\sigma(-\beta)=-\sigma(\beta)$. The subgroup $H$ of permutations
satisfying those two conditions consists of eight elements : in cycle notation,
$$
\begin{array}{lcl}
H &=\lbrace& {\sf id},(\alpha,-\alpha)(\beta,-\beta),\\
 & & (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta), \\
 & & (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta),  \\
 & & (\alpha,-\alpha), (\beta,-\beta) \rbrace
\end{array} 
$$
There exactly three transitive subgroups of $H$, namely $H$ itself and
$$
\begin{array}{lcl}
H_1 &=&\lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta),
 (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta)  \rbrace \\
H_2 &=& \lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta),
 (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta)  \rbrace 
 \end{array}
$$
In case (a), $(\alpha\beta)^2=b$ is a square in $K$, so 
$\gamma_1=\alpha\beta$ is in $K$. Now if $\tau_1=(\alpha,-\beta,-\alpha,\beta)$, we have
$\tau_1(\gamma_1)=-\gamma_1$, so $\tau_1\not\in G$ and this forces
$G=H_1$.
In case (b), $(\alpha\beta(\alpha^2-\beta^2))^2=b(a^2-4b)$ is a square in $K$, so 
$\gamma_2=\alpha\beta(\alpha^2-\beta^2)$ is in $K$. Now if $\tau_2=(\alpha,\beta)(-\alpha,-\beta)$, we have
$\tau_2(\gamma_2)=-\gamma_2$, so $\tau_2\not\in G$ and this forces
$G=H_2$.
Finally, in case (c) we have $\gamma_1\not\in K$ and $\gamma_2\not\in K$. By the
fundamental theorem of Galois theory, $\gamma_1$ is not fixed by all
the elements of $G$, so $G\neq H_2$. Similarly  $\gamma_2$ is not fixed by all
the elements of $G$, so $G\neq H_1$. The only possibility left is then
$G=H$.
