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In Dummit's Abstract Algebra on page65 exercise16(c), there is a proposition "If $G=\langle x\rangle$ is a cyclic group of order $n\geq1$ then a subgroup $H$ is maximal if and only if $H=\langle x^p\rangle$ for some prime $p$ dividing $n$ ."

But I find that for a composite $k$ not dividing $n$ , $\langle x^k\rangle$ may be also maximal although $k$ is neither prime nor divding $n$ .

For instance, let $G=\langle x\rangle$ be a cyclic group of order $10$ ,then $|\langle x^6\rangle|=\frac{10}{\gcd(6,10)}=5$ implies that $\langle x^6\rangle$ is a maximal subgroup of $G$ since there does not exist a composite $c$ s.t. $|\langle x^6\rangle|=5c=10=|\langle x\rangle|$ .

So whether this propositon is correct? Or if we should restrict $k$ to be the smallest integer in $\langle x^k\rangle$ ?

Quay Chern
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  • The keyword is some, i.e. $\exists.$ $H$ is maximal iff there exists a prime $p$ dividing $n$ s.t. $H=\langle x^p\rangle.$ The generators $H$ are then all the $x^{pm}$'s with $m\in[1,n/p)$ coprime to $n/p,$ so you are right: $p$ is the smallest of such exponents $pm.$ In your example, $p=2$ and $m=1,2,3,4.$ – Anne Bauval Apr 29 '23 at 05:08
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    Does this answer your question? Characterizing maximal subgroups of cyclic groups. – Anne Bauval Apr 29 '23 at 05:11
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    Oh, I get your point. The notion "for some" connects with the fact that $\langle x^6\rangle$ and $\langle x^2\rangle$ are equivalant when $n=10$ . Thank you! And your link also does help. – Quay Chern Apr 29 '23 at 06:20

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I want to point out that both your example and the proposition in Dummit and Foote are correct. Note that in your example, $\langle x^6\rangle$ can also be written as $\langle x^2\rangle$ (i.e. $x^2$ and $x^6$ generate the same subgroup).

Stultus sum
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