# A group with a finitely generated subgroup of finite index

$G$ 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.
• For (2), see also math.stackexchange.com/questions/573050/…. – lhf Nov 30 '13 at 14:55
• (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. – Ran Kashtan Nov 30 '13 at 15:09

## 1 Answer

(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 $[G:H]=n$.

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