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This is from Pinter's Book of Abstract Algebra Chapter 11, Exercise D5.

Let $n=\operatorname{ord}(a)$.

I think I can prove the $\Rightarrow$ direction: since $a^r$ generates $\langle a \rangle$ iff $\gcd(r, n)=1$, hence $a=b^k$ generates $\langle b \rangle$. Since $a$ generates $\langle a\rangle$ and $\langle b \rangle$, the equality holds.

But I have trouble proving the $\Leftarrow$ direction. What I can see so far:

$\langle a\rangle\subseteq\langle b\rangle$ because $a = b^k\tag 1$

$n\mid\operatorname{ord}(b)\tag 2$ because the order of cyclic subgroup $A$ of cyclic group $B$ divides the order of cyclic group B. $\operatorname{ord}(b)\mid k n\tag 3$ because $b^{kn}=a^k=e$

Because of $(1)$, we know that $\langle a \rangle=\langle b\rangle$ iff $a$ and $b$ have the same order, so it feels like I just need to tighten $(2)$ and $(3)$, but I am stuck. Any help will be appreciated.

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  • $\begingroup$ Related $(1)$ $\endgroup$
    – PinkyWay
    Jun 5, 2020 at 6:20
  • $\begingroup$ Related $(2)$. $\endgroup$
    – PinkyWay
    Jun 5, 2020 at 6:29
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    $\begingroup$ Wow, thanks for the edits! $\endgroup$
    – Risan
    Jun 5, 2020 at 7:47
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    $\begingroup$ I put the links in case some other idea occurs to you. 😀Those might be helpful in future. $\endgroup$
    – PinkyWay
    Jun 5, 2020 at 8:01
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    $\begingroup$ Also, if you want "$\mid$" for divides , use \mid and, if you want $\nmid$, use \nmid. 😀 $\endgroup$
    – PinkyWay
    Jun 5, 2020 at 8:12

1 Answer 1

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Looks like it is false, which I overlooked because I was under the assumption that it is true.

Counter example: Consider the cyclic subgroup of $\mathbb{Z}_{10}$. Let's say $a = 2$ and $b = 1$. We know that $a = b^2$, $\operatorname{ord}(a) = 5$, and $\gcd(2, 5) = 1$. However, $\langle 1 \rangle = \mathbb{Z}_{10}$ but $\langle 2 \rangle$ consists of all even numbers of $Z_{10}$. Hence $\langle a \rangle \neq \langle b \rangle$.

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  • $\begingroup$ I recently asked this same question. Can I assume that the question from Pinter's book has a typo since the converse doesn't hold? I.e. it is not the case that $\langle a\rangle = \langle b\rangle$ whenever $a = b^k$ and the order of $a$ and $k$ are relatively prime? $\endgroup$ Oct 23, 2023 at 22:41

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