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This a little exercise mentioned in my book:

Consider a subgroup $G \le S$ of index $2$. $G$ is normal in $S$ and has a unique coset $S\backslash G$ that is not itself.

Suppose that we have the property that all elements of $S$ belong to $G$ or none of them belong to $G$ (and thus to $S\backslash G$).

Can we say that for each $x,y \in S$, the product $xy$ is an element of $G$?

My attempt:

It is clear that if both $x$ and $y$ belong to $G$, that $xy \in G$. But what if $x,y \in S\backslash G$? I think $xy \in S\backslash G$, but that would be too easy and I feel like this is a trick question.

Am I right? Thanks.

MyWorld
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    "all elements of $S$ belong to $G$ or none of them belong to $G$" What exactly does this mean? Does it mean "$G = S$ or $G=\varnothing$"? Does it mean that for any element, it's either in $G$ or not in $G$? Or does it mean something else? – Arthur May 08 '19 at 14:23
  • It's distracting that $S$ is not the subgroup. – Randall May 08 '19 at 14:25
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    $x\in S\setminus G$, then $x^{-1}\in S\setminus G$, but $xx^{-1} = 1\in G$. – Groups May 08 '19 at 14:26

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