-1
$\begingroup$

$G$ is a group, and $H$ is a subgroup of $G$ with index $[G:H]=n$. Prove or disprove the following:

  1. If $H$ is a finitely generated group then G is a finitely-generated group.
  2. If $a\in G$ then $a^n\in H$.
  3. If $a \in G$ then $H\cap \{a,a^2,...,a^n\}\ne \emptyset$.

Progress

  1. If $G$ is finite, then $G$ is a finitely-generated group (as all finite groups). If $G$ isn't finite, then due to the finite index, I conclude that $H=G$, and then $G$ is a finitely-generated group.
  2. Thank you lhf for the answer.
  3. I need help here.
$\endgroup$
  • $\begingroup$ For (2), see also math.stackexchange.com/questions/573050/…. $\endgroup$ – lhf Nov 30 '13 at 14:55
  • $\begingroup$ (1) If G is final, then G is a finitely-generated group (as all final groups). If G isn't final, then due to the final index, I conclude that H=G, and then G is a finitely-generated group. (2) Thank you lhf for the answer. (3) I need help here. $\endgroup$ – Ran Kashtan Nov 30 '13 at 15:09
1
$\begingroup$

(1) Let $H=\langle K\rangle, |K|<\infty$ and $\{x_1,\ldots,x_n\}$ be representatives of the cosets of $H$. Then $G=\langle K\cup\{x_1,\ldots,x_n\}\rangle$ is finitely generated.

(3) Let $H\cap \{a,a^2,...,a^n\}= \emptyset$. Then all of $a,a^2,...,a^n$ are contained in different cosets, and every coset $a^iH$ is different from $H$. So there are $\ge n+1$ cosets, this contradicts $[G:H]=n$.

$\endgroup$
  • $\begingroup$ @James Mitchell Thank you very much. $\endgroup$ – Boris Novikov Nov 30 '13 at 20:50
  • $\begingroup$ You're welcome! $\endgroup$ – James Mitchell Dec 1 '13 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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