Proving A^n = e for Finite G with Identity e

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SUMMARY

The discussion centers on proving that for any element \( a \) in a finite group \( G \) with identity \( e \), there exists a positive integer \( n \) such that \( a^n = e \). The reasoning involves recognizing that if \( a^n \) were not equal to \( e \), the sequence of powers \( a, a^2, a^3, \ldots \) would eventually repeat due to the finiteness of \( G \), leading to \( a^k = a^j \) for some integers \( j < k \). This implies \( a^{k-j} = e \), confirming that every element in a finite group has finite order.

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  • Understanding of group theory concepts, specifically finite groups.
  • Familiarity with the identity element in group theory.
  • Knowledge of the concept of order of an element in a group.
  • Basic mathematical reasoning and proof techniques.
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  • Study the properties of finite groups in group theory.
  • Learn about the concept of element order in groups.
  • Explore the implications of Lagrange's theorem in finite groups.
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Homework Statement


Show that if a belongs to G where G is finite with identity e, then there exists a positive n such that a^n = e


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The Attempt at a Solution



Someone tell me if i am reasoning correctly.

Suppose a^n is not equal to e.

By finiteness of G, a^n = b for some b belonging to G. Then b^t = (a^n)^t is not equal to e for any positive t. so G is infinite? i don't know i am stuck.
 
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since a is in G, then the elements a, a^2, a^3, a^4, ... are also in G, but G is finite, so a^k = a^j for some j, k, say j < k, so a^(k-j) = e, set n = k - j, so a^n = e and n > 0.

realize what you proved, you proved that every element of a finite group has finite order.
 
In your attempt, you assume b^t =/= e for all t, which isn't a great assumption, considering you're supposed to prove that (in this case given b) there IS t for which b^t=e
 

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