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Revision difference : Proof: The orbit of any element of a group is a subgroup |
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| Following is a proof that, if $G$ is a group and $g \in G$, then $\langle g \rangle \le G$. Here $\langle g \rangle$ is the orbit of $g$ and is defined as |
Following is a proof that, if $G$ is a group and $g \in G$, then $\langle g \rangle \le G$. |
| $$\langle g \rangle=\{g^n : n\in\Z\}$$ |
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| Since $g \in \langle g \rangle$, then $\langle g \rangle$ is nonempty. |
Since $g \in \langle g \rangle$, then $\langle g \rangle$ is nonempty. |
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| Let $a,b \in \langle g \rangle$. Then there exist $x,y \in {\mathbb Z}$ such that $a=g^x$ and $b=g^y$. Since $ab^{-1}=g^x(g^y)^{-1}=g^xg^{-y}=g^{x-y} \in \langle g \rangle$, it follows that $\langle g \rangle \le G$. |
Let $a,b \in \langle g \rangle$. Then there exist $x,y \in {\mathbb Z}$ such that $a=g^x$ and $b=g^y$. Since $ab^{-1}=g^x(g^y)^{-1}=g^xg^{-y}=g^{x-y} \in \langle g \rangle$, it follows that $\langle g \rangle \le G$. |
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