Let $G$ be a group acting on a set $A$. Let $N$ be a non-trivial normal subgroup of $G$. Suppose that $S$ is an $N$-invariant set, i.e. $n \cdot s \in S$ for all $s \in S$, $n \in N$. Must $S$ be $G$-invariant?

I suspect the answer is yes (for example, it is true in the situation of a Galois group acting on a Galois field extension). I've been playing around with it for a while and haven't been able to get anything.

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    $\begingroup$ Sorry N-invariant means N(S)=S? $\endgroup$ – mathpadawan Dec 6 '18 at 23:36
  • $\begingroup$ @mathnoob Yes, I edited it for clarity. $\endgroup$ – vukov Dec 6 '18 at 23:37
  • $\begingroup$ What if $N$ is the trivial normal subgroup $\{e\}$? $\endgroup$ – Catalin Zara Dec 6 '18 at 23:39
  • $\begingroup$ @CatalinZara You're right, I meant to require non-trivial. $\endgroup$ – vukov Dec 6 '18 at 23:42

Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S=\{eN\}$. Then letting $G\cdot T=\{g\cdot t\mid g\in G, t\in T\}$ for any $T\subseteq A$, we have in particular $N\cdot S=S$ but $G\cdot S=G/N\neq S$.


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