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I have a question, in a sense, about how asymmetric left and right cosets can be when dealing with an infinite, non-normal subgroup $K$ of a (non-abelian) group $G$. Specifically, my question is whether there can exist a subgroup $K$ of a group $G$ such that for some (note crucially: just one, not all) $a \in G$, we find that $aK \subset Ka$ but $Ka \nsubseteq aK$?

We already know from the basic theory of normal subgroups that this is impossible if $K$ is finite or if it is normal. We further know that if (and only if) $Ka \subseteq aK$ and $Ka^{-1} \subseteq a^{-1}K$, then $aK = Ka$ as well, since in fact $Ka^{-1} \subseteq a^{-1}K \Leftrightarrow aK \subseteq Ka$. Thus, any example of my desired situation would have this property for $a$ but not for $a^{-1}$.

I appreciate any input/references, thanks.

MathNeophyte
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I construct an example. Let $G$ be the group $\langle x,y \mid y x = x^2 y \rangle$. Note that $G$ is non-abelian. The relations move the $y$ on the left to the right.

Let $K$ be the subgroup $\langle x \rangle$. Then $yK \subseteq Ky$. But $xy \in Ky$ and $\not\in yK$.

Robert Shore
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Functor
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  • When you say $G$ is the group $\langle x, y \mid yx = x^2y \rangle$ does this mean it is the group of all strings of powers of $x$ and $y$, where both generators have infinite order and interact according to the given rule? I'm not really familiar with the notation beyond cyclic groups, where I've seen $\langle x \rangle$ used for finite order cyclic groups as well.

    The crux of the argument then seems to just be that $yx^k = x^{2k}y$ for all $k \in \mathbb{Z}$, and supposing $xy = yx^k = x^{2k}y \Leftrightarrow x = x^{2k}$ contradicts the infinite order of $x$, correct?

     – MathNeophyte Apr 27 '24 at 01:21
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    Yes, but your description is not precise. The notation includes two parts: "free generation" and "relations". The free generation $\langle x,y \rangle$ really means they are free and have no relations. The only relations that can be used to reduce are $x x^{-1} = 1$ and $y y^{-1}=1$. For $\langle x,y \ |\ yx=x^2 y\rangle$, it adds a relation. But you should be careful that sometimes $x, y$ may have finite order. – Functor Apr 27 '24 at 01:54