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).

  • If $H$ is not equal to any $<s>$ then for each $<s>$ we have either some $h_s\in H $ \ $<s>$ or $H $ is a proper subset of $<s>$. – DanielWainfleet Oct 23 at 5:22
up vote 9 down vote accepted

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 $G$.

  • Do you mean $\neg (\forall x \in G, \langle x\rangle = H)$? The you've worded it, it sounds like you're saying $\forall x \in G, \langle x\rangle\ne H$. – Acccumulation Oct 22 at 20:46
  • @Acccumulation The additional hypothesis (for the proof by contradiction) is that $H$ is not cyclic; thus, for every $x\in H$, $\langle x\rangle\ne H$; on the other hand, $\langle x\rangle\ne H$ when $x\notin H$; thus $\langle x\rangle\ne H$ for every $x\in G$. This doesn't lead to an easy contradiction, though. – egreg Oct 22 at 20:51

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 $G$

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 $H$.

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".

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.