Proof: The orbit of any element of a group is a subgroup
Following is a proof that, if is a group and , then . Here is the orbit of and is defined as
Since , then is nonempty.
Let . Then there exist such that and . Since , it follows that .
Title | Proof: The orbit of any element of a group is a subgroup |
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Canonical name | ProofTheOrbitOfAnyElementOfAGroupIsASubgroup |
Date of creation | 2013-03-22 13:30:58 |
Last modified on | 2013-03-22 13:30:58 |
Owner | drini (3) |
Last modified by | drini (3) |
Numerical id | 6 |
Author | drini (3) |
Entry type | Proof |
Classification | msc 20A05 |
Related topic | Group |
Related topic | Subgroup |
Related topic | ProofThatEveryGroupOfPrimeOrderIsCyclic |
Related topic | ProofOfTheConverseOfLagrangesTheoremForCyclicGroups |
Defines | orbit |