Tate Cohomology of Squares Let $L | \mathbb{Q}_p$ be a finite Galois extension with Galois group $G$. What is known about the group $H^{-1} (G, L^\times / (L^\times)^p)$?
I'm particularly interested in the case of a non-abelian wildly ramified extensions $L | \mathbb{Q}_p$. Even more precisely, I need to show that $H^{-1} (A_4, L^\times / (L^\times)^2) = 0$, where $L | \mathbb{Q}_2$ denotes the unique extension having Galois group $A_4$. (I am confident this statement holds due to computations using MAGMA.)
This is how far I've come yet: There is an element $x \in L^\times$ such that $N_{L | K} (x) = - 1$ and the statement above is, since $H^{-1}( A_4, L^\times) \cong \mathbb{Z} / 2 \mathbb{Z}$, equivalent to saying that $x^2 \neq 0$ in $H^{-1} (A_4, L^\times)$. Using the exact sequence
$$ 1 \to H^{-1} (A_4, L^\times) \stackrel{\cdot 2}{\to} H^{-1}(A_4, (L^\times)^2) \to \{ \pm 1\} \to 1 $$
obtained from the long exact sequence induced by $L^\times \stackrel{\cdot 2}{\to} (L^\times)^2$ this is equvialent to $H^{-1} (A_4, (L^\times)^2)$ being cyclic. 
However, I do not know how to proceed from here as I feel I've literally squeezed out every information one could get from related long exact sequences in cohomology. Can you suggest what else I could try or where I could look for getting inspiration?
 A: I don't think that Tate cohomology is useful here, at least because we don't control a priori $H^{-1} (G, L^*)$, whereas $H^1 (G, L^*)=0$ by Hilbert's $90$. Let us first fix the general setting: $L/K$ is a finite Galois extension of fields of characteristic $\neq 2$, and you want to classify the quadratic extensions $L'/L$ s.t. $L'/K$ is Galois. NB: this is the problem exposed in your comment. The calculation of $H^{-1}(G,L^*/{L^*}^2)$ could also be carried on following the method explained below, but I don't see the motivation.
Since $L$ contains $W_2:=(\pm 1)$, every quadratic extension $L'/L$ is of the form $L'=L(\sqrt a)$, $a\in L^*, a\notin {L^*}^2$; actually, $L'$ depends only on the class $\bar a\in X_L:=L^*/{L^*}^2$. An easy classical argument shows that $L'/K$ is Galois iff $a/s(a)\in {L^*}^2$ for all $s\in G$; in other words our extensions $L'$ are classified by the group of invariants ${X_L}^G$, and our task is the description of ${X_L}^G$ in cohomological terms. This will be done by studying the $G$-cohomology of the two tautological exact sequences $1\to {L^*}^2\to L^* \to X_L\to 1$ and $1\to W_2\to  L^*\to {L^*}^2\to 1$.
The first exact sequence gives $...\to X_K \to {X_L}^G \to H^1(G, {L^*}^2) \to H^1(G, L^*)=0$. Besides, denoting by $\epsilon$ the natural map $X_K \to {X_L}^G$, it is obvious  that Ker $\epsilon =(K^* \cap {L^*}^2)/{K^*}^2$, which  is no other than Hom $(G, W_2)$ by Kummer theory. Summarizing, we have an exact sequence  (1) $0\to X_K $/Hom $(G, W_2)$ $\to {X_L}^G \to H^1(G, {L^*}^2)\to 0$ , and it remains to describe $H^1(G, {L^*}^2)$.
The second exact sequence gives (2) $0\to H^1(G, {L^*}^2) \to H^2(G, W_2)\to H^2(G, L^*)\to ...$ Here the $G$-action on $W_2$ is trivial, hence $H^2(G, W_2)$ depends only on the structure of $G$, and given $G$, it can be dismantled using general cohomological tools. Try for example $G=A_4$. As for the relative Brauer group $H^2(G, L^*)$  attached to the extension $L/K$ , it depends on the field $K$. If for instance $K=\mathbf Q_p$ (or more generally a $p$-adic local field), local CFT tells us that $H^2(G, L^*)$ is cyclic of order $[L:K]$; since $H^2(G, W_2)$ is killed by $2$, its image will be $0$ or $\mathbf Z/2$, the distinction between the two cases depending on normic computations.
NB : the above study carries on word for word when replacing $2$ by any prime $p$, under the hypothesis that $K$ has characteristic $\neq p$ and contains the group $W_p$ of $p$-th roots of $1$.
Complement. You added a question whether all the Galois extensions $L'/\mathbf Q_2$ thus obtained are split. This can be answered by manipulating inflation-restriction, as I said, to reduce to more amenable quotients of $G$. Let us fix some notations : $k=\mathbf Q_2$ for simplicity, $G=Gal(L/k)=$ $A_4, H=C_2 \times C_2 =$ the $2$-Sylow subgroup of $A_4, F=$ the fixed field of $H, G^{ab}=G/H=C_3$. It is classically known that $H^2(G, W_2)=C_2$ and $H^2(H, W_2)=C_2 \times C_2$ . 
Let us first rewrite the exact sequence (2) as (3)  $0\to H^1(G, {L^*}^2) \to H^2(G, W_2)\to C\to 0$, where $C=C_2$ or $(1)$ denotes the image of $H^2(G, W_2)$ in the relative Brauer group $H^2(G, L^*)$. Replacing $k$ and $G$ by $F$ and $H$, we obtain a similar exact sequence (4) $0\to H^1(H, {L^*}^2) \to H^2(H, W_2)\to C'  \to 0$, with $C'=C_2$ or $(1)$, and the restriction map yields a commutative diagram with exact lines (3) and (4) and vertical connecting maps denoted $res_1 , res_2 , res_3$ [unfortunately I don't know how to type commutative diagrams].
1) By local CFT, the rightmost $res_3$ between the two relative Brauer groups is just multiplication by $3=[F:k]$, hence an isomorphism. Consider $res_1$, which takes place in the inf-res exact sequence $0\to H^1 (G^{ab},F^* \cap {L^*}^2)\to H^1(G,{L^*}^2)\to {H^1(H,{L^*}^2)}^G \to H^2 (G^{ab},F^* \cap {L^*}^2)$. Since $G^{ab}$ is cyclic, the $G^{ab}$-cohomology of the exact sequence $1\to {F^*}^2 \to F^* \cap {L^*}^2 \to (F^* \cap {L^*}^2)/ {F^*}^2 \to 1$ reads ${\hat H}^0((F^* \cap {L^*}^2)/ {F^*}^2)\to H^1({F^*}^2)\to {H^1(F^*\cap {L^*}^2)}^G \to H^1((F^* \cap {L^*}^2)/ {F^*}^2)$ (we omitted to write $G^{ab}$). Because $2$ and $3$ are coprime, the first and last term vanish to give an isomorphism $H^1({F^*}^2)\cong {H^1(F^*\cap {L^*}^2)}^G$. But we have seen above that $H^1({F^*}^2)$ describes the quadratic extensions $M/F$ which are Galois over $k$, and these are necessarily split because $2$ and $3$ are coprime. Hence $res_1$ is injective, and consequently so is $res_2$. Applying the snake lemma to the commutative diagram we get an exact sequence $0\to$ Coker $res_1 \to C_2 \to$ Coker $res_3\to 0$ . Denote by $c,c'$ the orders of $C, C'$, and by $h^1(G), h^1(H)$ the orders of $H^1(G, {L^*}^2),H^1(H, {L^*}^2)$. A simple diagram chase gives $h^1(H)/h^1(G)=2c/c'$ .
2) It remains to compute $H^1(H, {L^*}^2)$, or rather its image in $H^2(H, W_2)$, which classifies the quadratic extensions $L''/L$ which are Galois over $F$. Their Galois groups, described by the classes of $H^2(H, W_2)$, are of three different types : $C_2 \times C_2, D_8$ (dihedral) or $H_8$ (quaternionic). The first type is split abelian, so we need to study only the two non abelian types using techniques of the embedding problem. Let $L=F(\sqrt \alpha, \sqrt \beta)$. It is easy to show that $L/F$ can be embedded into a $D_8$-extension, cyclic over $F(\sqrt \alpha\beta)$ iff $\alpha$ is a norm from $F(\sqrt \beta)$. In our case here, this normic condition can be expressed in terms of the local Hilbert symbol $<.,.>$ attached to $F$; it simply reads $<\alpha, \beta>=0$ (in additive notation). The embeddability into an $H_8$-extension is more complicated; a necessary and sufficient condition reads $<\alpha,-1>+<\beta,-1>+<\alpha, \beta>=0$ (Fröhlich, (1985)). To conclude we need to know an explicit description of $L/F$, hence my previous question about how you got yo $A_4$-extension of $\mathbf Q_2$. But I forgot about the Fröhlich condition ! The prime $2$ is always a mess.
A: In addition to Nguyen Quang Do's excellent and comprehensive answer I want to show how one can answer my complementary question on whether all quadratic extensions $L' | L$ such that $L' | \mathbb{Q}_2$ is Galois define in fact a split Galois extension $L' | \mathbb{Q}_2$. (This is equivalent to the question on the vanishing of $H^{-1} (A_4, L^\times / (L^\times)^2)$ raised in my original post.)
Observe that any such quadratic extension $L' | L$ yielding a Galois extension $L' | \mathbb{Q}_2$ is in fact a central extension of $L | \mathbb{Q}_2$, hence the Galois group of $L' | \mathbb{Q}_2$ is $\mathbb{Z}/2\mathbb{Z} \times A_4$ or $SL_2 (\mathbb{F}_3)$ (these are the only central extensions of $A_4$). Suppose $G_{L ' | \mathbb{Q}_2} = SL_2 (\mathbb{F}_3)$, then $L' | F$, where $F | \mathbb{Q}_2$ is the unique subextension of $L | \mathbb{Q}_2$ of degree 3, is a Galois extension with Galois group $Q$ (the quaternion group). It therefore suffices to show that $L | F$ can not be embedded into a extension with Galois group $Q$. To do so, we will be using Fröhlich's Condition, which asserts that this is the case if and only if
$$ \langle a, b \rangle \cdot \langle a, - 1\rangle \cdot \langle b, - 1 \rangle = - 1, \qquad  \quad (\ast)$$
where $\langle \cdot, \cdot \rangle$ denotes the Hilbert symbol and $L = F(\sqrt{a}, \sqrt{b})$ (see (7.7) in Fröhlich, A.: Orthogonal representations of Galois groups, Stiefel-Whitney classes and Hasse-Witt invariants, 1985). 
We will therefore now first determine $a$ and $b$.
It is easy to see that $F| \mathbb{Q}_2$ is unramified, hence $F = \mathbb{Q}_2 (\alpha)$, where $\alpha \in F^\times$ is a root of $X^3  + X + 1$. Denote by $\varphi$ the Frobenius of $F | \mathbb{Q}_2$. One now checks that $\{2, -1, 1 + 2\alpha,  1 + 2 \varphi (\alpha), 5\}$ is an $\mathbb{F}_2$-basis of $F^\times / (F^\times)^2)$ and that
$$ N_{F | \mathbb{Q}_2} (1 + 2\alpha) = N_{F | \mathbb{Q}_2} ( 1 + 2 \varphi(\alpha)) = - 3 \equiv 5 \mod (F^\times)^2. $$
The action of $G_{F | \mathbb{Q}_2}$ on $F^\times / (F^\times)^2$ is therefore given by 
$$ \varphi (1 + 2 \alpha) = 1 + 2 \varphi (\alpha), \qquad \varphi (1 + 2 \varphi (\alpha)) = 5 (1 + 2 \alpha)(1 + 2 \varphi (\alpha)) $$
and fixing all other basis vectors. By writing down explicitely all orbits of this action one sees that the only $G_{F | \mathbb{Q}_2}$-stable subgroup of order 4 of $F^\times / (F^\times)^2$ on which $G_{F | \mathbb{Q}_2}$ acts non-trivially is 
$$\{ 1, 5(1 + 2 \alpha), 5(1 + 2 \varphi(\alpha)), (1 + 2 \alpha)(1 + 2 \varphi(\alpha)) \}.$$
Hence $L = F \left (\sqrt{5(1 + 2 \alpha)}, \sqrt{5(1 + 2 \varphi(\alpha)})\right)$ by Kummer Theory. A similar discussion (looking for stable subgroups $U$ with non-trivial action on the quotient of $F^\times / (F^\times)^2$ by $U$) gives that the norm group of $L | F$ is 
$$ N_{L | F} (L^\times) = \langle 2 \rangle \times \mu_7 \times \langle -1, U^{(2)}_F \rangle, $$
where $\mu_7 \subseteq F^\times$ denotes the $7$-th roots of unity. One can now, using $\langle a, -1 \rangle = \langle -1, a\rangle^{-1}$, check that $(\ast)$ holds. 
