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 $G$.
Best Answer
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.