In the unit group of a real biquadratic field, what is the index of the product of the unit groups of its three quadratic subfields?
Is the index 1 if discriminant of these three subfields are always 1 mod 4?
What can we say about the index when the biquadratic field is not real?
Ad 1: The index is easily seen to be finite, and the details will depend on how, exactly, the Galois group acts on the large unit group---and this in turn correlates to some extent with how it acts on ideal classes. (A buzzword is multiplicative Galois module structure. These structures become reasonably nice when one is at liberty to pass to $S$-units, i.e. to make additional elements invertible, killing class groups and ramification and other obstacles. In this sense, the global units are as far away from niceness as you can get.)
Ad 2: Not necessarily. $\mathbb{Q}\left(\sqrt{5},\sqrt{29},\sqrt{145}\right)$ provides a counterexample:
The quadratic subfields can be defined, respectively, using PARI-friendly variables, by $y_1^2-y_1-1=0$ with fundamental unit $\varepsilon_1=y_1$, and $y_2^2-5y_2-1=0$ with fundamental unit $\varepsilon_2=y_2$, and $y_3^2-y_3-36=0$ with fundamental unit $\varepsilon_3=2y_3+11$. The compositum is generated by a root of $x^4-17x^2+36$. The automorphism $\sigma:x\mapsto -x$ fixes the third subfield, and $x \mapsto -17/6\,x + 1/6\,x^3$ fixes the first. A fundamental system of units upstairs is $$\eta_1=-\frac{1}{2}-\frac{11}{12}x + \frac{1}{12}x^3 = -\varepsilon_1,$$ $$\eta_2=-\frac{5}{2}-\frac{23}{12}x + \frac{1}{12}x^3 = -\varepsilon_2,$$ $$\eta_3=-1-\frac{1}{3}x+\frac{1}{2}x^2+\frac{1}{6}x^3$$ and we recover $\varepsilon_3$ as $-\eta_3^2\eta_1/\eta_2$ after observing that $\sigma(\eta_3)=\eta_1\eta_3/\eta_2$. Thus $$-\varepsilon_3 \varepsilon_2/\varepsilon_1$$ is a square in the large unit group.
Ad 3: The complex (CM) case is a bit simpler in one respect since the unit rank is $1$ and the cyclic group generated by the fundamental unit of the real subfield is of finite index in the whole unit group. On the other hand, nontrivial roots of unity intervene. (Consider $\mathbb{Q}\left(\sqrt{2},\sqrt{-1},\sqrt{-2}\right)$ where the eighth roots of unity do not come from the subfields.) As in all CM field extensions, you can identify complex conjugation $\sigma$ with the abstract automorphism of order $2$ which fixes the real subfield. For any unit $\eta$ of the biquadratic field, $\sigma(\eta)/\eta$ is the inverse of its complex conjugate, thus all its embeddings into $\mathbb{C}$ lie on the unit circle; since it is an algebraic integer, it must be a root of unity by a classic lemma of Kronecker. And $\eta\sigma(\eta)$ is a unit of the real subfield. These observations constrain what can happen; you just need to be extra careful with fields containing the fourth or sixth roots of unity.
Systematic results might require a study of the cyclotomic units involved, depending on what kind of result you're after. See Washington's Cyclotomic Fields for an introduction.
$\def\Z{\mathbf{Z}}$ $\def\eps{\epsilon}$
Observe that if a unit $\epsilon \in L$ lives in a quadratic subfield, then $N(\epsilon) = + 1$. So the index of the units is positive whenever $L$ has a unit of norm $-1$. If $L = \mathbf{Q}(\sqrt{5},\sqrt{13})$, then
$$\epsilon = \frac{9 + \sqrt{5} + \sqrt{13} + \sqrt{65}}{4}$$
is a root of the polynomial $x^4 - 9 x^3 + 20 x^2 - 7 x - 1$, from which one can see that $\epsilon$ is a unit of norm $-1$.
More generally, the index will always be a (small, bounded) power of two (proof, the units of $L$ inject into the units of the subfields under the composition of the three norm maps, whereas the subgroup of units coming from subfields under this map has index $8$). This problem was studied by Kubota. He proved a relationship between the index of this unit group and the ratio of the class number of $L$ to the product of the class numbers of the quadratic subfields. (explicitly, the index is $4h/h_1h_2h_3$.)
There is a nice discussion of this in Section 7 of Maria Stadnik's paper here:
http://arxiv.org/pdf/1202.3475v1.pdf,
from which one learns the following nice generalization of the computation above (see Proposition 7.6): if $p$ and $q$ are primes such that $p \equiv q \equiv 1 \mod 4$ and the Legendre symbol $\left(\frac{p}{q}\right) = -1$, then $\mathbf{Q}(\sqrt{p},\sqrt{q})$ always has a unit of norm $-1$, and so the index is always $>1$ (in fact, it will always be $2$ in this case.)
Let me also quote the following from the paper above concerning the work of Kubota: if the fundamental units of the subfields of $L$ are $\epsilon_1$, $\epsilon_2$, and $\epsilon_3$, then Kubota "completely classified the possible structures into one of seven types." As an example, "if every subfield has a unit of norm $-1$, then the units of $L$ are either of the two forms $\{\epsilon_1,\epsilon_2,\epsilon_3\}$ or $\{\epsilon_1,\epsilon_2,\sqrt{\epsilon_1, \epsilon_2, \epsilon_3}\}$", the latter occurring if and only if $L$ has a unit of norm $-1$.
In the imaginary case, if $\epsilon$ is the fundamental unit of $L$ (the unit rank is one) and $K$ is the quadratic subfield, then $N_{L/K}(\epsilon)$ is a unit of $K$. On the other hand, if $\eta$ is the fundamental unit of $K$, then, thought of as an element of $L$, $N_{L/K}(\eta) = \eta^2$. Hence, up to roots of unity in $L$, either $\eta$ is the fundamental unit of $L$ or is a square of the fundamental unit. If the latter occurs, then $\eta \zeta$ is a square in $L$ for a root of unity $\zeta \in L$. Both cases may happen:
In $L = \mathbf{Q}(\sqrt{-1},\sqrt{5})$, the unit group is $\mu_4 \oplus \Z$, where $\Z$ is generated by the fundamental unit of $\mathbf{Q}(\sqrt{5})$.
In $L = \mathbf{Q}(\sqrt{-1},\sqrt{6})$, the unit group is $\mu_4 \oplus \Z$, where $\Z$ is generated by $\displaystyle{\eps = \displaystyle{\frac{1-\sqrt{-1}}{2}} \cdot (2 + \sqrt{6})},$
and $\eps^2 = \zeta \eta$, where $\zeta^4 = 1$ and $\eta = (5 + 2 \sqrt{6})$ is the fundamental unit of $\mathbf{Q}(\sqrt{6})$. (So the relevant indices in these two cases are two and four respectively.)
