What is wrong with this fake proof that subgroup of a cyclic group is cyclic? Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $\langle r \rangle = H$. If some $s$ in $G$ is not in $H$ then $\langle s \rangle \neq H$. So for all $s$ in $G$, $\langle 
s \rangle \neq H$ Thus for all $s$ in $G$ there exists an $h\in H$ such that $h \notin \langle s \rangle$. But this is a contradiction with the fact that $a$ generates $G$.
I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).    
 A: No $s\in G$ such that $\langle s\rangle=H$ is clear.
But then you can't follow with for all $s\in G$ there's $h\in H$ such that $h\notin\langle s\rangle$ because $H$ could be a subset of $\langle s\rangle$ even though $s\notin G$.
Which, incidentally, is what happens if $s$ is a generator of $G$
A: You are correct in proving that, for every $x\in G$, $\langle x\rangle\ne H$. But this doesn't imply that, for every $x\in G$, there is $h\in H$ such that $h\notin\langle x\rangle$.
For instance, this is false as soon as $H\ne G$ and $x$ is a generator of $G$.
A: It is correct that $\forall s \in G, \langle s\rangle \ne H$, but this does not imply $\exists h \in H, h \notin \langle s\rangle$, because $\langle s\rangle$ may be a proper superset of $H$.
A: 
If some $s$ in $G$ is not in $H$ then $\langle s \rangle \neq H$. So for all $s$ in $G$, $\langle s \rangle \neq H$ 

That doesn't make any sense. Your saying "$(s \not \in H \rightarrow \langle s \rangle \neq H)\rightarrow (\forall s, $$\langle s \rangle \neq H)$". That's like saying "if an animal doesn't have a tail, it's not a dog, therefore no animals are dogs".
