Order of a^k (Group Theory)

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Today we will revise some basic Group Theory. Let G be a group and a\in G. Assume that a has finite order n. Find the order of a^k where k is an integer.

Answer: \displaystyle|a^k|=\frac{n}{(n,k)}, where (n,k)=\gcd(n,k).

Proof:

Our strategy is to prove that m=\frac{n}{(n,k)} is the least smallest integer such that (a^k)^m=1.

Now, we have \displaystyle a^{k\cdot\frac{n}{(n,k)}}=(a^n)^{\frac{k}{(n,k)}}=1. Note that k/(n,k) is an integer and thus a valid power.

Suppose to the contrary there exists c<\frac{n}{(n,k)} such that a^{kc}=1.

Since a has finite order n, we have n\mid kc, which leads to \displaystyle\frac{n}{(n,k)}\mid\frac{k}{(n,k)}\cdot c. Note that \frac{n}{(n,k)} and \frac{k}{(n,k)} are relatively prime.

Thus \frac{n}{(n,k)}\mid c, which implies that \frac{n}{(n,k)}\leq c which is a contradiction. This proves our result. 🙂

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