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If $G$ is a cyclic group of order $n$ and $k\mid n$, then $G$ has exactly one subgroup of order $k$.

My attempt: $k\in \Bbb{N}$ is implicitly assumed. Let $G=\langle a\rangle$ and $|G|=|a|=n$. Since $k\mid n$, we have $n=kq$, for some $q\in \Bbb{N}$. Let $H=\langle a^q\rangle$. By theorem 4(vii) section 1.3, $|H|=\frac{n}{q}=k$. Suppose $\exists H\leq G$ such that $|H|=k$. By theorem 5 section 1.3, $H=\langle a^m\rangle$ where $m$ is least positive integer such that $a^m\in H$. Since $|H|=k$, we have $$|H|=|\langle a^m\rangle |=\frac{n}{m}=\frac{kq}{m}=k.$$ Which implies $q=m$. So $H=\langle a^q\rangle$. Thus $\exists ! H\leq G$ such that $|H|=k$. Is my proof correct?

Edit: I just realize, my proof is wrong. $|\langle a^m\rangle |=\frac{n}{m}$ if $m$ satisfy $m\mid n$ condition. In general $|\langle a^m\rangle |=\frac{n}{(n,m)}$. I have found a way to fix my proof. By theorem 6 section 1.3, $H= \langle a^m\rangle = \langle a^{(n,m)}\rangle$. So $|H|=\frac{n}{(n,m)}=\frac{kq}{(n,m)} =k$. Which implies $q=(n,m)$. Thus $H= \langle a^q\rangle$.

user264745
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    It might help to [edit] the question to include theorem 4(vii) section 1.3. – Shaun Apr 01 '23 at 18:53
  • @Shaun To be honest, I could have done same thing without citing those theorem. I cited those theorem for future references. – user264745 Apr 01 '23 at 18:54
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    Does this answer your question? https://math.stackexchange.com/q/4329046/977780 – Sourav Ghosh Apr 01 '23 at 19:00
  • @SouravGhosh ahh… kind of, I mean question is the same, but answers are bit different. I used “subgroup of a cyclic group is a cyclic group” fact. You took slightly longer route. – user264745 Apr 01 '23 at 19:08
  • I think your answer (in the original post, before edited) is correct. $|<a^q>|=k$ is true. A brief proof of $|<a^q>|=k$: Since $|<a^q>|$ is finite, $<a^q>={a^{qi}| i\in\mathbb{Z}_+}$. It is obvious that $e=a^0, a^q, a^{2q}, \cdots, a^{(k-1)q}$ are distinct. And for $i\ge n$, suppose $i=jk+l$, $0\le l\le k-1$, then $iq=jkq+lq=jn+lq\Longrightarrow a^{iq}=a^{lq}$. Note that $l\in[0,\ k-1]$. So we conclude $e=a^0, a^q, a^{2q}, \cdots, a^{(k-1)q}$ are all the distinct elements of $<a^q>$. – Asigan Apr 03 '23 at 00:10
  • Your writing is very confusing, "Let $H=\langle a^q\rangle$. By theorem 4(vii) section 1.3, $|H|=\frac{n}{q}=k$. Suppose $\exists H\leq G$ such that $|H|=k$." Why do you say "Suppose exist H"? Do you mean to make an assumption and finally derive a contradiction? Maybe you can write "suppose exsit another subgroup $H_2$ which has the same order $|H_2|=k$", Since you have already constructed $H=\langle a^q\rangle$ and $|H|=\frac{n}{q}=k$, next you need to prove the uniqueness of your contructed subgroup $H=\langle a^q\rangle$ – MathFail Apr 03 '23 at 09:58
  • @Asigan your conclusion $|a^q|=|\langle a^q\rangle |=k$ is correct. Your proof could be more precise. Theorem 4(vii) section 1.3 states “$|a|=n$ and for each $k\in \Bbb{N}$ such that $k\mid n$, $a^k=n/k$”. Sketch of proof: show $(a^k)^{n/k}=e$ and $\nexists 0\lt r\lt n/k$ such that $a^r=e$. Proposition 5 section 2.5 of Dummit and Foote states “If $|a|= n\lt \infty$ and $k\in \Bbb{Z}-{0}$, then $|a^k|=n/(n,k)$”. Sketch of proof: show $\langle a^k\rangle = \langle a^{(n,k)}\rangle$ and use theorem 4(vii) to conclude $|a_k|=n/(n,k)$, since $(n,k)\mid n$ by definition of gcd. – user264745 Apr 03 '23 at 10:23
  • @Asigan The reason why $|H|=\frac{n}{q}$ is correct because $q\mid n$. – user264745 Apr 03 '23 at 10:28
  • @MathFail first I showed existence part after that uniqueness. Proving uniqueness using “suppose exsit another subgroup $H_2$ which has the same order $|H_2|=k$” statement uses contradiction argument. I’m not a anti-contradiction proof person. Both (your and mine) are same essentially same way to show uniqueness. – user264745 Apr 03 '23 at 10:39
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    @user264745 Yes, what I have done is simply giving another proof, independent of theorem 4(vii) section 1.3. The theorem is much stronger. – Asigan Apr 03 '23 at 12:43

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