I'm having some trouble understanding the proof of the following theorem

A subgroup of a cyclic group is cyclic

I will list each step of the proof in my textbook and indicate the places that I'm confused and hopefully somewhere out there can clear some things up for me.


Let $G$ be a cyclic group generated by "$a$" and let $H$ be a subgroup of $G$. If $H = {\{e\}}$, then $H = \langle e\rangle $ is cyclic. If $H \neq \space {\{e\}}$, then $a^n \in H$ for some $n \in \mathbb{Z}^{+}$. Let $m$ be the smallest integer in $\mathbb{Z}^{+}$ such that $a^m \in H$.

We claim that $c = a^m$ generates $H$; that is,

$$H = \langle a^m\rangle = \langle c\rangle.$$

We must show that every $b \in H$ is a power of $c$. Since $b \in H$ and $H \leq G$, we have $b = a^n$ for some $n$. Find $q$ and $r$ such that

$$ n = mq + r \space \space \space \space for \space \space \space 0 \leq r < m,$$

Alright this is the first part in the proof where I start to get confused. Where does the division algorithm come from and why are we using it? The proof continues as follows:

$$a^n = a^{mq + r} = (a^m)^q \cdot a^r,$$


$$ a^r = (a^m)^{-q} \cdot a^n.$$

Now since $a^n \in H$, $a^m \in H$, and $H$ is a group, both $(a^m)^{-q}$ and $a^n$ are in $H$. Thus

$$(a^m)^{-q} \cdot a^n \in H; \space \space \space \text{that is,} \space \space a^r \in H.$$

This is another point at which I’m a little confused. What exactly about $a^n$ and $a^m$ being elements of $H$ allows us to accept that $(a^m)^{-q}$ and $a^{n}$ are in $H$? The proof continues:

Since $m$ was the smallest positive integer such that $a^m \in H$ and $0$ $\leq r$ $< m$, we must have $r = 0$. Thus $n = qm$ and

$$b \space = \space a^n \space = \space (a^m)^q \space = \space c^q,$$

so $b$ is a power of $c.$

This final step is confusing as well, but I think its just because of the previous parts I was confused about. Any help in understanding this proof would be greatly appreciated

  • 2
    $\begingroup$ "Where does the division algorithm come from?" is a very vague question. In this cases, there is a deeper theorem involved, that is hidden because you are talking about groups rather than the (possibly later topic) rings, name that the integers are something called a "principal ideal ring." The division algorithm is deeply entwined with this property. $\endgroup$ Feb 5, 2013 at 17:19
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    $\begingroup$ What textbook is this? This is a nice proof. $\endgroup$ Jun 1, 2017 at 3:57
  • $\begingroup$ Could you advise which textbook this was adapted from? Thanks. $\endgroup$ Sep 11, 2018 at 10:48
  • 1
    $\begingroup$ For the questions under the main question regarding where the proof came from, the book where this proof is from is: A First Course in Abstract Algebra by John B.Fraleigh. I know this does not answer the main question, whoever can add comments could move this there. $\endgroup$ Jan 24, 2020 at 18:59

4 Answers 4


For the first question, the appearance of the division algorithm is best explained by its usefulness in the rest of the proof. You might think of it because you want $n=qm$, as you need to show that $a^n$ is a power of $a^m$, but the best you can do at that point is say $n=qm+r$ and then attempt to prove $r=0$.

For the second question, as $a^m\in H$, we have $(a^m)^{-1}\in H$, as subgroups are closed under taking inverses, and then $(a^m)^{-q}=((a^m)^{-1})^q\in H$, as subgroups are closed under multiplication.

  • 1
    $\begingroup$ .@Matt Can you please explain how r =0 in this proof ....i havenot understood following two lines from proof.....< Since m was the smallest positive integer such that am∈H and 0 ≤r <m, we must have r=0. Thus n=qm > thanks $\endgroup$
    – Taylor Ted
    Apr 3, 2015 at 10:37
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    $\begingroup$ You assume $m$ is the smallest positive integer such that $a^m\in H$, but then find that $a^r\in H$, and $r<m$. The only way this is possible is if $r=0$. $\endgroup$
    – mdp
    Apr 3, 2015 at 10:50

First, the use of the division algorithm is for the sake of utility, as it provides a basis for the structure of the proof. Here, we want to relate $n$ with $m$, that is, we want to show that $n$ must be a multiple of $m$, but we start with the fact that what we know of any $n$, then given any positive integer $m$, by the division algorithm, there exist unique integers integers $q$ and $r$, where $0\leq r\lt m$ such that $n = mq + r$.

Essentially that means for any $n$, if we divide by a positive integer $m$, we have a unique integer quotient $q$ and a unique integer remainder $r$ where $0\leq r \lt m$. So to show that $m$ is a multiple of $n$ (no remainder), we want to show that $n = qm + 0$. I.e. we want $n=qm$, in order to show that $a^n$ is a power of $a^m$, but at most we can start with the fact that $n=qm+r$. The objective then is to prove that $r=0$.

For the second question, since $a^m\in H$, it follows $(a^m)^{-1}\in H$, since subgroups are closed under taking inverses. Then, since $a^m, (a^m)^{-1} \in H$, $(a^m)^{-q}=((a^m)^{-1})^q\in H$, since subgroups are closed under multiplication.

Does this clarify matters any?


In the first answer given, at the last part, $a^r$ is a member of $H$, and $0\le r<m$. Since $a^r$ is a member of $H$ and $a^m$ is a member of $H$ where $m$ is the smallest integer s.t. $a^m$ belongs to $H$.

Thus $r>m$. However $r<m$, by Euclid's division lemma. Hence, due to this contradiction, $r=0$, and $n=qm$.


a nontrivial cyclic group is a group with a singleton generating set, and vice versa.

let H be a cyclic group, and suppose $K$ is a non-cyclic subgroup.

evidently $K$ is a proper subgroup of H and has no set of generators of cardinality less than 2.

choose a generator $h$ for H.

in examining generating sets for $K$, we may exclude any containing the identity.

and since $K$ is proper no generating set contains $h$

for $K$ choose a generating set $\mathfrak{K}$ which contains an element $h^m$ where $m \gt 1$ is minimal amongst all the powers of $h$ occurring in generating sets for K.

without loss of generality we may assume that for any $p \gt 1$ we have $h^{pm} \notin \mathfrak{K}$

since $|\mathfrak{K}| \ge 2$

$$ \exists n \gt m.h^n \in \mathfrak{K} $$ define $a \ge 1$ by $$ a = \max\{b|bm \lt n\} $$ it follows that: $$ 0 \lt n-am \lt m $$ but since $h^{n-am}=h^n (h^m)^{-a} \in K$ it follows that $\mathfrak{K}'=\mathfrak{K} \cup \{h^{n-am}\}$ is a generating set for $K$ contradicting the minimality in our choice of $m$


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