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.