Abstract Algebra: Every group has a cyclic subgroup I have to show that every group has a cyclic subgroup. I know what this means, and to me it is obvious, yet I am not sure how to formally write it.
I proved it directly, as follows:

Let $G$ be a group.
Let $g$ be an element in $G$.
Let $o(g) = k$, i.e., $g^k = e$, for some integer $k$,
so by definition, $\langle g\rangle = \{e, g, g^2, \dots, g^{k-1}\}$,
therefore, for some $g\in G$, the cyclic subgroup generated by $g$ is in fact a subgroup of $G$.

Would that be sufficient to prove the statement? Did I leave anything out, or should I mention anything else?
New version of the proof:

Let $G$ be a finite group.
If $G = \{e\}$, then $G = \langle e\rangle$, i.e., cyclic.
If $G\neq\{e\}$, then there exists $g\neq e$, for $g\in G$.
Let $m$ be minimum positive integer s.t. $g^m = e$.
Then $g, g^2, \dots, g^m = e$ are distinct elements in $G$ generated by $g$
So $e, g, g^2, \dots, g^{m-1}$ form a cyclic subgroup $\langle g\rangle$ in $G$.

 A: Your proof is incorrect. You take an element $g\in G$ and immediatelly claim that $o(g) = k$ for some integer $k$. However, you cannot claim that, as there are groups in which no element has finite order.
A: As already said, it is not true that an arbitrary group has always a finite cyclic subgroup: take $\Bbb Z$ under addition.
However, what is true, is that an arbitrary group $G$ has always a cyclic subgroup. Take an element $g\in G$ and consider the subgroup of $G$ generated by this element: $\langle g\rangle$.
You have now two cases: 
1)$\operatorname{ord}(g)$ is finite, say $N$; then
$$
\langle g\rangle=\{1_G,g,g^2,\dots,g^{N-1}\}=\{g^k\;:\;0\le k\le N-1\}
$$
for example you can take $G=S_n$ the symmetric group and $g=(1,2,\dots,n)$, or even $G=\Bbb Z_n$ the class of remainder modulo $n$, which is a group under addition, and take $g=1$. An interesting example is the one given by the group $GL_n(\Bbb R)$ which is an infinite group taking a idempotent matrix as $g$.
2)$\operatorname{ord}(g)=\infty$: the situation is as above, except the fact that the list of elements in $\langle g\rangle$ never ends, i.e.
$$
\langle g\rangle=\{1_G,g,g^2,\dots,g^{k},\dots\}=\{g^k\;:\;k\in\Bbb Z_{\ge 0}\}
$$
for example take $G=\Bbb Z$ with $g=1$ or even $GL_n(\Bbb R)$ as above with a non idempotent matrix as $g$.
