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Suppose H is an arbitrary subgroup of G and $H\neq \{e\}$. Let $a^{k_{0}}\in H$. Let $b=a^{t}$ be an arbitrary member of H such that $k_{0}\le t$. For some $q,r$ s.t. $t=k_{0}q+r$, then $a^{t}=a^{k_{0}q+r}=a^{k_{0}q}a^{r}$ for $0\le r\lt k_{0}$. Since $k_{0}\le t$, $r=0$. Thus, $a^{t}=a^{k_{0}q}$ and therefore H is cyclic. Since H was arbitrary, any subgroup of a cyclic group is also cyclic.

Is this proof correct? Any possible improvements welcome

Shaun
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Not Friedrich gauss
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    If $a^{k_0}=e$ so $a^{k_0 q}=e^q=e$, so unless your subgroup is the trivial one there is a mistake – Marco Nov 03 '23 at 22:50
  • @That was a mistake. Instead, i just stated that $a^{k_{o}}\in H$ without the condition. That should fix it. Thanks – Not Friedrich gauss Nov 03 '23 at 22:56
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    Alternative: wlog, $~H \neq {e}.~$ Let $~n~$ denote the smallest positive integer such that $~b = a^n \in H.~$ Then, prove that $~H = \langle b\rangle.$ – user2661923 Nov 03 '23 at 23:00
  • "Since $k_0<t$, $r=0$." Why should this be true? – Greg Martin Nov 03 '23 at 23:13
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  • Posts don't start on the title (just like letters don't start on the envelope). Make sure the question you are trying to ask is included in the post itself and not just in the title line. 2. The solution-verification tag description says: ""Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof that is in doubt, and why so.". Your post, right now, is just "Is my proof correct?" and as such is off topic.
    • – Arturo Magidin Nov 03 '23 at 23:19
  • @GregMartin Suppose $k_{0}\ge t$ If $k_{0}>t$ then $r>1$. However, $k_{0}=t$ then $r>1$ So in fact i should have denoted $k_{0}\le t$. Thanks – Not Friedrich gauss Nov 03 '23 at 23:21
  • @ArturoMagidin understood – Not Friedrich gauss Nov 03 '23 at 23:22
  • The step I identified in my earlier comment is (still) unjustified. – Greg Martin Nov 03 '23 at 23:49
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. Compare your proof to the linked common proofs. If any questions remain then ask questions in comments on these proofs. If that fails, then edit your question to identify precisely which step you question, and why so. – Bill Dubuque Nov 04 '23 at 07:23
  • @GregMartin Well $k_{0}\le t$ implies that its the smallest element in the set – Not Friedrich gauss Nov 04 '23 at 22:01