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[parent] Proof: The orbit of any element of a group is a subgroup (Proof)

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

\begin{displaymath}\langle g \rangle=\{g^n : n\in\mathbbmss{Z}\}\end{displaymath}

Since $g \in \langle g \rangle$, then $\langle g \rangle$ is nonempty.

Let $a,b \in \langle g \rangle$. Then there exist $x,y \in {\mathbbmss{Z}}$ 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 $\langle g \rangle \le G$.



"Proof: The orbit of any element of a group is a subgroup" is owned by drini. [ full author list (2) | owner history (2) ]
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See Also: group, subgroup, proof that every group of prime order is cyclic, proof of the converse of Lagrange's theorem for finite cyclic groups

Also defines:  orbit
Keywords:  group, subgroup

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Cross-references: group
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This is version 3 of Proof: The orbit of any element of a group is a subgroup, born on 2003-03-12, modified 2003-06-04.
Object id is 4102, canonical name is ProofThatGInGImpliesThatLangleGRangleLeG.
Accessed 5489 times total.

Classification:
AMS MSC: 20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)
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