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?
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 $H\subseteq \langle a^d\rangle$.
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, $H = \langle \phi (x) \rangle$.