## 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. ðŸ™‚

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