# A^n = e

1. Nov 26, 2007

### rsa58

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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 dunno i am stuck.

2. Nov 26, 2007

### 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.

3. Nov 26, 2007

### Office_Shredder

Staff Emeritus
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