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Problem 50 (The Inverse In A Finite Group Is A Power Of The Element)
Let $G$ be finite group with $a\in G$. Prove that there exists a positive integer $k$ such that $a^k=a^{-1}$.
Hint: Consider the sequence $a, a^2, a^3, a^4, \ldots$. Why must you eventually have a repeated element? So if $a^i=a^j$ for some $j>i$, how can you use this to find $a^{-1}$?
The following pages link to this page.
- Problem.FiniteSubgroupTest
- Problem.TheInverseInAFiniteGroupIsAPowerOfTheElement
- Schedule.20161014
- Schedule.20161017
- Schedule.20171013
- Schedule.20171016
- Schedule.20171020
- Schedule.20171023
- Schedule.AllProblems
- Solution.TheInverseInAFiniteGroupIsAPowerOfTheElementChristian
- Solution.TheInverseInAFiniteGroupIsAPowerOfTheElementJason
- Solution.TheInverseInAFiniteGroupIsAPowerOfTheElementTyler