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As with the element-wise conjugation action, for $H$ and $K$ subgroups of $G$ in the same $G\text{-orbit}$ under this action, say that $H$ and $K$ are conjugate.

I know that $Gs_o = \lbrace gs_o\ :\ g ∈ G\rbrace$ can be one orbit for G-orbits, but I am not just getting what "same $g\text{-orbit}$" means.

If I am somehow mistaken, can anyone show me where I got wrong in the meaning of G-orbits and relate it to the comprehension of the quote?

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up vote 3 down vote accepted

Let $\mathcal{S}$ be the set of subgroups of $G$. For $H \in \mathcal{S}$ and $g \in G$, we denote $gHg^{-1}$ by $g.H$. Then $G$ acts on $\mathcal{S}$ by $H \rightarrow g.H$. If $H, K \in \mathcal{S}$ are in the same $G$-orbit in this action, we say $H$ and $K$ are conjugate.

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The $G$-orbit (under conjugation) of a subgroup $H$ is the set of subgroups of the form $gHg^{-1}=\{ghg^{-1}:h\in H\}$ for some $g\in G$. When we say that two subgroups $H$ and $K$ are in the same $G$-orbit, we mean that there is some $g\in G$ such that $gHg^{-1}=K$.

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