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Does every non prime order finite group have at least one non trivial proper cyclic subgroup?

Let $a\neq e$ be an element of $G$. Then $\langle a \rangle$ is always a subgroup of $G$, rather it is a cyclic subgroup of $G$. But my doubt arises, what will guarantee us that we shall always have at least one such proper subgroups generated by the elements of $G.$

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By Cauchy theorem, if $p\mid n$ and order of $G$ is $n$ then there exists an element of order $p$ in $G$ call it $g$ then $\langle g\rangle$ a cyclic subgroup of order $p$. Examples: $\langle r\rangle \leq D_8$ and $\langle 2\rangle \leq \mathbb{Z}/12\mathbb{Z}$

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  • $\begingroup$ Cauchy's theorem is way overkill here. $\endgroup$ – Tobias Kildetoft Feb 12 '20 at 18:57
  • $\begingroup$ Thanks, actually Cauchy's theorem together with the fundamental theorem of arithmetic makes the whole thing clear. $\endgroup$ – sukanta das Feb 12 '20 at 19:08
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Hint Pick any $a \neq e$.

Look at $<a>$. If this is strictly smaller than $G$ you are done.

If $<a>=G$ then $ord(a)=|G|$ which is composite. Therefore $ord(a)=mn$ for some $m,n >1$. Show that $<a^m>$ is a proper cyclic subgroup.

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  • $\begingroup$ Yes, nice argument. Thank you. $\endgroup$ – sukanta das Feb 12 '20 at 19:11

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