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Lemma $1.92$ in Rotman's textbook (Advanced Modern Algebra, second edition) states,

Let $G = \langle a \rangle$ be a cyclic group.

(i) Every subgroup $S$ of $G$ is cyclic.

(ii) If $|G|=n$, then $G$ has a unique subgroup of order $d$ for each divisor $d$ of $n$.

I understand how every subgroup must be cyclic and that there must be a subgroup for each divisor of $d$. But how is that subgroup unique? I'm having trouble understanding this intuitively. For example, if we look at the cyclic subgroup $\Bbb{7}$, we know that there are $6$ elements of order $7$. So we have six different cyclic subgroups of order $7$, right?

Thanks in advance.

Community
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    Keith Conrad has some wonderful notes on this very topic. See Theorem 3.6 for your specific question. http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicgp.pdf – Alex Wertheim Jun 03 '13 at 18:32
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    This article might help. See "Subgroups of cyclic groups" that relate to the one you are talking about. Hope this helps. – NasuSama Jun 03 '13 at 18:33
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    @Artus, I will point out that you have some mistakes in your thinking. By "The cyclic subgroup 7", I assume you mean a cyclic group of order 7, say, the additive group $\mathbb{Z}/7\mathbb{Z}$. In this case, take two nonzero elements, say $1$, and $2$. Write out by hand the subgroups that these two elements generate. Are they different? – Alex Wertheim Jun 03 '13 at 18:36
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    This Question was closed as a duplicate of one about subgroups of cyclic groups being cyclic. However the present Question focuses instead on the existence and uniqueness of subgroups of a finite cyclic group of order $n$ given a {positive) divisor $d$ of $n$ to be the order of the (cyclic) subgroups. I'm voting to reopen. – hardmath Aug 04 '23 at 00:39

5 Answers5

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Let $d$ be a divisor of $n=|G|$. Consider $H=\{ x \in G : x^d =1 \}$. Then $H$ is a subgroup of $G$ and $H$ contains all elements of $G$ that have order $d$ (among others).

If $K$ is a subgroup of $G$ of order $d$, then $K$ is cyclic, generated by an element of order $d$. Hence, $K\subseteq H$.

On the other hand, $x\in H$ iff $x=g^k$ with $0\le k < n$ and $g^{kd}=1$, where $g$ is a generator of $G$. Hence, $kd=nt$ and so $k=(n/d) t$. The restriction $0\le k<n$ implies $0\le t<d$, and so $H$ has exactly $d$ elements. Therefore, $K=H$.

lhf
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    I like this proof a lot. Just thought I'd say :) – user1729 Jul 27 '13 at 19:59
  • The inequality $0 \leq t <d$ do give a bound for $t$ but how can we conclude from this that for every integral $t$ satisfying the bound there exists a $k$? –  Nov 11 '14 at 12:02
  • @user170039, it follows directly from $k=(n/d) t$, doesn't it? Note that $n/d$ is an integer. – lhf Nov 11 '14 at 12:16
  • How does it follow exactly? I can't figure out... – Meitar Jan 21 '15 at 20:11
  • I can't see from where t came from... –  Mar 16 '16 at 17:17
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    @Alnitak, if $g^m=1$ then $m$ is a multiple of $n$, the order of $g$. – lhf Mar 16 '16 at 18:11
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    @lhf The restriction $0\le k<n$ implies $0\le t<d$, and so $H$ has exactly $d$ elements. Can you please clarify how you made a conclusion about $H$? I don't see the connection between upper bound on $t$ and size of $H$. – False Promise Jan 04 '17 at 19:08
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    @FalsePromise, $d$ possible values of $t$ iff $d$ possible values of $k$. – lhf Jan 04 '17 at 20:56
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Suppose $\langle a \rangle$ has order $n$. If $d \mid n$, then $a^{n/d}$ has order $d$. Any subgroup of $\langle a \rangle$ is of the form $\langle a^k \rangle$ for some $k \mid n$. Hence if a subgroup has order $d \mid n$, it must be $\langle a^{n/d} \rangle$.

You are right that there are $6$ elements of order $7$ in a cyclic group of order $7$, but these all generate the same cyclic subgroup.

Mikko Korhonen
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To help you understand where you're going wrong, why not try writing out these "six different subgroups": if $G$ is a cyclic group of order $7$, and $a$ is a generator of $G$, then

$$\begin{array}{c|c} \mathsf{\text{Subgroup of }}G\mathsf{\text{ generated by}} & \mathsf{\text{consists of}}\\\hline a \strut & a,\; a^2,\;a^3,\; a^4,\; a^5,\;a^6,\;a^7=e\\\hline a^2 \strut& \\\hline \vdots \strut&\\\hline a^6\strut & \\\hline \end{array}$$

Zev Chonoles
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Hint $\ $ The key idea for $\,\Bbb Z/n\,$ is the same as the proof for $\,\Bbb Z\!:\,$ since a subgroup S is closed under subtraction, every element of S is a multiple of the least element $> 0$ (using representatives $\ge 0$).

Bill Dubuque
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Key Ideas
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Suppose $|G| = n$ and $G = \langle a \rangle = \{e,a,\ldots,a^{n-1}\}$. If $d$ divides $n$, then the subgroup $\langle a^{n/d} \rangle$ has order $d$. Conversely, suppose that $H = \langle g \rangle$ is a subgroup of order $d$ (we know it's cyclic by the first part). Now we know that $g = a^{k}$ for some $1 \leq k \leq n-1$. If $H$ has order $d$, can $k$ be anything other than $n/d$?

Suugaku
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