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Problem 62 ($\langle a^k\rangle = \langle a^{\gcd(k,|a|)}\rangle$)

Let $a$ be an element of order $n$ and let $k\in\mathbb{N}$. Prove that $\langle a^k\rangle = \langle a^{\gcd(k,n)}\rangle$.

Solution

Let $a\in G$ have finite order $n$. Let $k\in \mathbb{Z}$. Let $d=\gcd(k,n)$. We will show that $\langle a^k\rangle = \langle a^{\gcd(k,n)}\rangle$, which is the same as showing $\langle a^k\rangle = \langle a^d\rangle$ . First we'll prove $\langle a^k\rangle \subseteq \langle a^{d}\rangle$. Let $x\in \langle a^k\rangle$. This means $x=(a^k)^p$ for some integer $p$. Note that $d$ is a divisor of $k$, as $d$ is the greasted common divisor. This means we can write $k=dq$ for some integer $q$. Substition gives us $$x=(a^k)^p=a^{kp} = a^{dqp}=(a^{d})^{qp},$$ which shows that $x\in \langle a^d\rangle$ as desired. This proves that $\langle a^k\rangle \subseteq \langle a^{d}\rangle$.

We now prove that $\langle a^d\rangle \subseteq \langle a^{k}\rangle$. First we use the GCD theorem to obtain $s$ and $t$ such that $d=sk+tn$. Now let $x\in \langle a^d\rangle$. This means $x=(a^d)^m$ for some integer $m$. We now compute $$ %\begin{eqnarray} x=(a^d)^m=a^{dm}=a^{(sk+tn)m}=a^{skm+tnm}=(a^{k})^{sm}(a^{n})^{tm}=(a^{k})^{sm}(e)^{tm}=(a^{k})^{sm}, %\end{eqnarray} $$ which shows that $x\in \langle a^{k}\rangle$, and completes the proof that $\langle a^d\rangle \subseteq \langle a^{k}\rangle$.

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