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Let $G=(g)$ be a finite cyclic group generated by $g$ with $|G|=n$, and let $d \in \mathbb{N}$ with $d|n$, then an unique subgroup $H$ of $G$ with $|H|=d$ exists.

Proof of existence: $\exists m \in \mathbb{N}: md=N$. Since $G=(g)$ we have $g^k\neq g^l\ \forall 0< k\neq l\leq N$ and $g^k=g^{k+nl}\ \forall k,l \in \mathbb{N}$.

Hence $g^{mk} \neq g^{ml}\ \forall 0< k\neq l \leq d$ and $g^{mk}=g^{m(k+nl)}\ \forall k,l \in \mathbb{N}$. Therefore $H=(g^m)=\{g^{m1},...,g^{md}=e\}$ is a (cyclic) subgroup of $G$ with $|H|=d$.

Is this ok?

Proof of uniqueness: Let $H_1,H_2$ be two subgroups with order $d$. As subgroups of a cyclic group $H_1,H_2$ are also cyclic. Set $k_1,k_2$ to be the smallest positive exponents so that $g^{k_1} \in H_1, g^{k_2} \in H_2$ respectively.

I do not know how I can continue from there to proof uniqueness..any help? Or is the approach wrong?

gulokol
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  • Prove that every subgroup is generated by $g^k$ for some $k \mid n$. Now what is the order of $\langle g^k \rangle$? What does this tell you about how many subgroups of order $d$ there can be? – Dustan Levenstein Jan 07 '14 at 17:44
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    possible duplicate subgroups of cyclic groups and their order – amWhy Jan 07 '14 at 17:45
  • Use the surjective homomorphism $\varphi\colon\mathbb{Z}\to G$ given by $\varphi(k)=g^k$. Then the answer easily follows from the homomorphism theorem – egreg Jan 07 '14 at 18:23
  • @amWhy The question you linked is much less general than this one – egreg Jan 07 '14 at 18:27
  • @egreg I simply thought it might help give insight. Note, I have not voted to close, so it is not currently in the "close queue." – amWhy Jan 07 '14 at 18:37
  • @egreg: As I thought to have shown the existence already, I am only left struggling with proofing the uniqueness. I tried to consider two cyclic subgroups of the same order, and then pick generating elements with minimal positive exponent. Then I considered the group homomomorphism $g^{k_1m} \mapsto g^{k_2m}$...Or how should I bring in the homomorphism? I do not get it..sorry.. – gulokol Jan 07 '14 at 18:54
  • For uniqueness, I believe you have to go to quotient groups. That is, suppose $H$ is any subgroup of order $d$, and use the quotient $G/H$ to show that $g^{n/d} \in H$. – Andreas Caranti Jan 07 '14 at 19:57

2 Answers2

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Consider the surjective homomorphism $$\varphi\colon\mathbb{Z}\to G,\qquad \varphi(k)=g^k$$ that has $\ker\varphi=n\mathbb{Z}$, where $g$ is a generator of $G$ and $|G|=n$.

If $H$ is a subgroup of $G$ with $|H|=d$, $d\mid n$, then we can compose $\varphi$ with the canonical projection to $G/H$, getting $$ \pi\circ\varphi\colon\mathbb{Z}\to G/H $$ that has kernel equal to $m\mathbb{Z}$, where $md=n$, because $m\mathbb{Z}$ is the only subgroup of $\mathbb{Z}$ having index $m$ and $|G/H|=m=n/d$. Therefore $$ \varphi^{-1}(H)=m\mathbb{Z} $$ so $H=\varphi(m\mathbb{Z})$, thus proving uniqueness.

egreg
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Here is another solution to this old problem:

As every finite cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n:=\mathbb{Z}/n\mathbb{Z}$ with the sum of integers modulo $n$, it is suffices to show this for $G=\mathbb{Z}_n$

Suppose $n=kp$, for $p,k\in\mathbb{N}$. If $p=1$ or $p=n$ then the corresponding subgroups of order $p$ are $(0)$ and $\mathbb{Z}_n$. for $1<p<n$ consider $H_p:=(0,p,2p,\ldots,(k-1)p)$. This is clearly a subgroup for order $k$ for if $$ap\equiv bp\mod(kp)$$ then $a\equiv b\mod k$ which means that the elements in $H_p$ do not repeat.

As for uniqueness, suppose $H'<\mathbb{Z}_n$ with $|H'|=k$. Let $p'$ be the smallest positive integer in $H'\cap\{1,1,\ldots,n-1\}$. Then $H'=\{0,p',2p',\ldots,(k-1)p'\}$ and $p'k=pk=n$. That is, $p=p'$.

Mittens
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