Cancellation of irreducibility for Galois conjugates Motivation: Take an algebraic number $\lambda$. In my research, I've stumbled upon the question in which cases the expression $\sum_{\sigma \in S} \sigma(\lambda)$, where $S$ is a subset of the field embeddings of $K=\mathbb{Q}(\lambda)$, can be 'less irrational' then $\lambda$ itself.
Question: Take a finite Galois extension $L/\mathbb{Q}$ with Galois group $G$, let $H \leq G$ be a subgroup and let $K$ be the field fixed by $H$. Assume that $K$ has no proper subfields except for $\mathbb{Q}$.
Consider a subset $S$ of $G/H$ and consider the $\mathbb{Q}$-linear vector space homomorphism
$$
f_S: K \to L, x \mapsto \sum_{ \sigma \in S} \sigma(x).
$$
Is it true that this map is injective unless $S=G/H$?
.
.
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Outdated Earlier version without the assumption that $K$ has no subfields (I leave this up only so Will's reply still makes sense) If $SH$ is a subgroup of $G$ strictly bigger than $H$, this map will not be injective because $f_S$ takes values in the subfield of $K$ fixed by $SH$. Intuitively, it seems likely that $SH$ being (the coset of) a subgroup is the only case in which this 'reduction of irrationality' can happen. Is this true, i.e. is $f_S$ injective under the condition that $SH$ is not the coset of a subgroup?
EDIT: As Will's example below shows below, one needs to exclude union of cosets, not only cosets. Indeed, if $U \subset G/H$ such that $UH$ is a subgroup, there is a linear relation which will carry over to any union of cosets of $UH$ as for $SH=UH \cup \sigma UH$,$f_S=f_U+\sigma \circ f_U$. Apologies, I updated the question.
 A: No.
Let $a, b \in \mathbb Q(i)$. Let $\alpha_1$ be a root of $x^3 + ax + b$.  Let $L$ be the field generated by $i, \alpha_1, \overline{\alpha}_1$.  Assume that $a,b$ are sufficiently general that $L$ has Galois group $S_3 \wr \mathbb Z/2$, i.e. $(S_3 \times S_3 ) \rtimes \mathbb Z/2$.
Let $K$ be the subfield generated by $\alpha_1 \overline{\alpha}_1$. Then $K$ is stabilized by $S_2 \wr \mathbb Z/2$, which is a maximal subgroup of index $9$, so $K$ has no proper subfields other than $\mathbb Q$.
Choose $\sigma_1,\sigma_2,\sigma_3$ embeddings which send $\alpha$ to the the three roots $\alpha_1,\alpha_2,\alpha_3$ of $f$ but preserve $\overline{\alpha}_1$ (possible by our assumption on the Galois group.
Then $$\sigma_1( \alpha_1 \overline{\alpha}_1) + \sigma_2( \alpha_1 \overline{\alpha}_1)+ \sigma_3( \alpha_1 \overline{\alpha}_1)=\alpha_1 \overline{\alpha}_1 + \alpha_2 \overline{\alpha}_1 + \alpha_3 \overline{\alpha}_1 = (\alpha_1 + \alpha_2 +\alpha_3) \overline{\alpha}_1 = 0 \overline{\alpha}_1=0$$
so $x\mapsto \sigma_1(x)+\sigma_2(x) + \sigma_3(x)$ is not injective.
A: The answer to your new question is still no.  I mentioned this problem (or rather, the group-theoretic reformulation given by Will Sawin) to my colleague Steve Humphries, and he found the following two examples.
Let $G$ be the 9th group of order 36 as indexed by the Magma code "SmallGroup(36,9)".  It is a Frobenius group with generators $a,b,c,d$ subject to the relations $$a^2 = b,
     b^2 = Id(G),
     c^3 = Id(G),
     d^3 = Id(G),
     c^a = c d^2,
     c^b = c^2,
     d^a = c^2 d^2,
     d^b = d^2.$$
Let $H=\{1,a,b,ab\}$, which is the subgroup generated by $a$ and $b$.  It is a maximal subgroup of index $9$ in $G$.
Let $s=Id(G) + c  d^2 + c^2  d$.  The three elements in the support of $s$ belong to distinct left cosets in $G/H$.
Let $t=\sum_{h\in H}h$ be the sum over the four elements of $H$, and let
$$
u=-4Id(G) - 7d - 7d^2 + 2c + 3c  d + b  d + b  d^2 + b  c  d^2 +
b  c^2
+ b  c^2  d + b  c^2  d^2 + a + a  d + a  d^2 + a  c  d^2 + a 
b  d +
a  b  d^2 + a  b  c^2  d^2.
$$
One can check (or have Magma check) that $tu\neq 0$, but $stu=0$.
For a bigger, but perhaps conceptually simpler, example work in the group $A_5$.  Take
$$s=(1, 4, 2, 5, 3) + (1, 5, 4) + (2, 4)(3, 5) + (1, 3, 5, 2, 4),$$ take
$$t=Id + (1, 2)(3, 4) + (3, 4, 5) + (1, 2)(4, 5) + (3, 5, 4) + (1, 2)(3, 5),$$ and take
$$
u=-4*Id - 6*(1, 2, 3, 4, 5) + 2*(1, 5, 4, 3, 2) - 6*(1, 3, 2) - 10*(1, 4,
2, 5, 3) +
     (1, 2, 4, 3, 5) + (1, 5, 4, 2, 3) + (1, 3)(2, 4) + (1, 4)(3, 5) +
(1, 2)(3, 4) + (1,
     3, 2, 5, 4) + (1, 2, 3, 5, 4) + (1, 5, 2) + (2, 3)(4, 5) + (1, 2,
4) + (1, 3)(4, 5) +
(1, 3, 2, 4, 5) + (2, 5, 3) + (1, 4, 2, 3, 5) + (3, 5, 4) + (1, 2, 3) +
(1, 3, 5, 4, 2) +
(1, 4)(2, 5) + (2, 3, 5) + (1, 2, 4, 5, 3) + (1, 5)(3, 4) + (1, 3, 5, 2,
4) + (2, 5)(3,
     4) + (1, 5)(2, 4).
$$
Then $tu\neq 0$ but $stu=0$.  Also notice that the support of $t$ is a maximal subgroup of $A_5$.  The elements in the support of $s$ belong to distinct left cosets.
