How to show every subgroup of a cyclic group is cyclic? I'm teaching a group theory course now, and I wanted to give my students a proof that every subgroup of a cyclic group is cyclic. The easiest way I could think to do this is to say that any cyclic group corresponds to a $\mathbb{Z}/n\mathbb{Z}$ and then (implicitly) use the first isomorphism theorem (which they haven't learned yet) to "reduce" it to the theorem that every subgroup of $\mathbb{Z}$ is of the form $m\mathbb{Z}$. The problem is that this latest statement is not simple at all (every subgroup is an ideal, and $\mathbb{Z}$ is a PID because $\mathbb{Z}$ is Euclidean).
This all seems way too convoluted for such an easy statement. What's the simplest most elementary proof that I can give?
 A: See Proposition 5 in these notes.
Note that part of this argument is left as an exercise for the reader.  It is an incarnation of the same argument that shows that a Euclidean domain is a PID, but one certainly doesn't need this terminology or any prior results.  Namely, one is supposed to show that -- upon identifying the elements of a cyclic group $Z_n$ with the integers $0,1,\ldots,n-1$ as usual -- if $H$ is a nontrivial subgroup of $Z_n$, then it is generated by its smallest nonzero element, say $k$.  Why?  Well, if not, there would exist an element $\ell$ of the subgroup which is not a multiple of $k$, and then by subtracting off a suitable multiple of $k$ -- to be completely explicit, let $a$ be the largest non-negative integer $a$ such that $\ell - ak$ is positive, and subtract $ak$ -- we get an element of the subgroup which is positive but smaller than $k$, contradiction.
The more explicitly ring-theoretic proof you have in mind is the one alluded to in the Remark at the bottom of page 2.
A: A simple way to present this and related results is to employ concrete fractional models of cyclic groups, i.e. $\rm\ C_n\: =\: \frac{1}n\mathbb Z\ \:(mod\ \mathbb Z)\ =\ \{ 0,\: 1/n,\: 2/n,\: \ldots,\: (n-1)/n\}\:.\ $ Suppose $\rm G$ is a nontrivial subgroup. Since $\rm G\:$ is closed under subtraction, every element of $\rm G\:$ must be an integer multiple of its least element $\rm\:g\:$ (else if $\rm\: h\:$ were the least nonmultiple, then $\rm\: h - g\:$ would be a smaller one). Thus $\rm\:G = \langle g \rangle\:$ is cyclic. $\ $ QED
A: If the original group is generated by $a$, and $d$ is the least positive integer such that $a^d$ is in the subgroup, then show using the division algorithm that $a^d$ generates the subgroup.

To elaborate, the division algorithm says:

If $m$ and $n$ are integers and $n$ is positive, then there are unique integers $q$ and $r$ such that $m=nq+r$ and $0\leq r\lt n$.

You could prove this by taking $r$ to be the least nonnegative integer expressible in the form $m-np$ as $p$ ranges over the integers, which then determines $q$.  (There are details here which I am omitting.)
Suppose $G$ is a cyclic group (finite or infinite) with generator $a$, i.e., $G=\langle a\rangle$.  Let $H$ be a subgroup of $G$.  Let $d$ be the smallest element of the set of positive integers $n$ such that $a^n$ is in $H$.  (Unless $G$ is infinite and $H$ is trivial, this exists.)  If $b$ is an arbitrary element of $H$, then because $b$ is in $G$, $b=a^m$ for some integer $m$.  Apply the division algorithm to write $m=dq+r$ for integers $q$ and $r$ with $0\leq r\lt d$.  Then $b=a^m=(a^d)^q\cdot a^r$.  Since $b$ is in $H$ and $a^d$ is in $H$, it follows that $a^r=a^m\cdot(a^d)^{-q}$ is in $H$.  If $r$ were positive, this would contradict the choice of $d$, so $r=0$, and $b=a^m=(a^d)^q$.  Since $b$ was arbitrary, this shows that $H\subseteq \langle a^d\rangle$.
A: It is enough to show that every maximal proper subgroup $H$ of $G$ is cyclic, by induction;
to start the induction, notice that if $G$ is simple, then there is nothing to prove.
So let $H\subset G$ be a maximal proper subgroup, so that $G/H$ is simple. There exists then a prime $p$ such that $G/H$ is isomorphic to $\mathbb Z/p\mathbb Z$ and then $pG\subset H$. From the exact sequence of abelian groups $$0\to H\to G\to G/H\to 0$$ we get the exact sequence $$0\to H/pG\to G/pG\to G/H\to 0.$$ Now this is an exact sequence of $\mathbb F_p$-vector spaces in an obvious way, and $G/H$ and $G/pG$ are vector spaces of dimension $1$ (for the first, this is obvious; for the second, it is enough to notice that $G/pG$ is cyclic as an abelian group and non-zero because it surjects onto $G/H$) It follows that $H/pG=0$, i.e., that $H=pG$. Since $G$ is cyclic, then $H$ is cyclic too.
A: If you accept Lagrange's theorem and
the existence of quotient groups, there's another
argument in the finite case.
Let $G=\langle a\rangle$ be a cyclic group of order $n$.
By Lagrange's theorem if $H$ is a subgroup of $G$ then $m=|H|$
is a factor of $n$, and if $m\mid n$ then $H_m=\langle a^{n/m}\rangle$
is a subgroup of order $m$. I claim $G$ has no other order $m$ subgroup.
The group $G/H$ has order $n/m$ so $x^{n/m}=e$ for all $x\in G/H$.
This translates to $x^{n/m}\in H$ for all $x\in G$ which means $H_m\le H$.
These have the same order so $H=H_m$.
A: Let's take $H$ a subgroup of $(Z/nZ,+)$. You can write $H=\{0, i_1,i_2, ..., i_r\}$ such that $0<i_1<i_2<i_3<...<i_r<n$. 
Now I claim that $i_2 = 2 i_1$. Since $H$ is a group, $i_2-i_1 \in H$, and $i_2 > i_2-i_1 > 0$. Obviously, $i_2 - i_1 = i_1$.
Now we prove that $i_3=3i_1$ : $i_3-i_1 \in H$ and $i_3 > i_3 - i_1 > 0$. So $i_3-i_1 = i_2$ or $i_3-i_1 = i_1$. In the second case, $i_3 = i_2$ which is false. So $i_3=i_2+i_1 =3i_1$.
I think now it's easy to prove by induction that $i_k = ki_1$ for all $1\leq k \leq r$.
A: How about:
Let G be a group. By definition, any quotient group H of G is realized by a surjective homomorphism $\phi: G \twoheadrightarrow H$. 
Now suppose $G = \langle x \rangle$. Then, since $\phi$ is surjective, $H = \langle \phi (x) \rangle$. 
