Motivation: Take an algebraic $\lambda$. 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. 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Outdated Earlier version without the assumption that $K$ has no subfields (I leave this up only so Will's reply still makes sense) If $SH$ is a subgroup of $G$ strictly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.