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

[rsa58]

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

Homework Equations

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.

 

[ircdan]

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.

 

[Office_Shredder]

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