# Prove that $G=\langle g_1,…,g_n\rangle$ if $G$ abelian.

Let $G$ a finite abelian group. Are there $g_1,...,g_m$ s.t. $$G=\langle g_1,...,g_m\rangle$$ where $g_i\neq g_j^m$ for all $m$ and all $i\neq j$ ? For cyclic group no problem, but I can't see such a group if $G$ is not cyclic, any example ?

• So one of the $g_i$s is the identity? – John Wayland Bales May 10 '16 at 21:06
• If $g_i = g_j^m$ for some $i, j, m$, then just remove $g_i$ from the presentation. Eventually you won't be able to do this and then you're done. – Qiaochu Yuan May 10 '16 at 21:07
• Klein four-group is abelian but not cyclic, and can be generated by any two of its non-identity elements. Also every non-identity element has order 2, so this property holds for the generator. – Pegah Pournajafi May 10 '16 at 21:08

Choose some arbitrary $g\in G$ for which there exists no $h\in G$ with $\langle{g}\rangle < \langle{h}\rangle$; such a $g$ must exist by the fact that $G$ is finite. (Otherwise, we'd have an infinite sequence of distinct subgroups of $G$.) Consider $G/\langle{g\rangle}$ and induct on $\#G$. (It may also be worth considering what fails in a divisible group, such as $\mathbb{Q}/\mathbb{Z}$.)