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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

We must show that $\langle a^k\rangle \subseteq \langle a^{\gcd(k, n)}\rangle$ and $\langle a^{\gcd(k, n)}\rangle \subseteq \langle a^k\rangle $.

First, we must show $\langle a^k\rangle \subseteq \langle a^{\gcd(k, n)}\rangle$. Let $b \in \langle a^k\rangle$. Then we know $(a^k)^m = b$ for some integer $m$. Let $m\in \mathbb{Z}$ such that $(a^k)^m = b$. Let $g = \gcd(n, k)$. Lastly, let $l = k/g$. Then we have that $k=gl$, so $(a^k)^m = (a^{gl})^m = b$. Using associativity, we get $b=(a^g)^{kl}$, so we have $b\in\langle a^g\rangle$. This means that $\langle a^k\rangle \subseteq \langle a^{\gcd(k, n)}\rangle$

Last, we must show $\langle a^{\gcd(k, n)}\rangle \subseteq \langle a^k\rangle $. Let $g = \gcd(k,n)$. Let $b\in \langle a^{g} \rangle$. Then we know that $b = (a^g)^m$ for some integer $m$. Let $m\in \mathbb{Z}$ such that $b = (a^g)^m$. By the greatest common divisor theorem, we know that $g=sk + tn$ for some $s, t \in \mathbb{Z}$, so $(a^g)^m = (a^{sk}a^{tn})^m$. Because $n$ is finite, we know $a^{tn} = a^0$, so $b = (a^{sk}e)^m$. Using associativity, we get $b = (a^k)^{sm} \in \langle a^k\rangle$. This means that $\langle a^{\gcd(k, n)}\rangle \subseteq \langle a^k\rangle $.

We have show that $\langle a^k\rangle \subseteq \langle a^{\gcd(k, n)}\rangle$ and $\langle a^{\gcd(k, n)}\rangle \subseteq \langle a^k\rangle $, so it must be that $\langle a^k\rangle = \langle a^{\gcd(k, n)}\rangle$

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