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 |
|---|---|
| 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 |