Let G be a group and let {$a_1$, ..., $a_n$} for $n\in \mathbb{N}$ be a finite subset, such that all $a_i$ mutually commute.
- As a set $\left\langle a_1, ..., a_n\right\rangle = \left\{a^{m_1}_1 ... a^{m_n}_n \mid m_i \in \mathbb{Z}\right\}$. Further, $\left\langle a_1, ..., a_n \right\rangle$ is abelian.
- $\langle e \rangle$ = {e}, where e is the neutral element of G.
- $\langle a\rangle$ = $\langle a^{-1}\rangle$ for any a ∈ G
I have been asked to prove this theorem, but I'm not sure where to begin as I don't really understand it. If someone would be able to explain this theorem or the way it can be used, that would be appreciated. Thank you :)
I assume that your definition of $\langle S\rangle$ is „the intersection of all subgroups of $G$ containing $S$ as a subset“. This definition immediately yields $S \subseteq \langle S\rangle$ with = holding precisely if $S$ already is a subgroup. In particular (2) follows immediately from this definition, as $\{e\}$ is the trivial subgroup.
For the rest note that by definition $S \subseteq H$ implies $\langle S\rangle \subseteq H$ for any subgroup $H$ of $G$. Hence for (3) it suffices to notice that necessarily $a^{-1} \in \langle a\rangle$ and $a=(a^{-1})^{-1} \in \langle a^{-1}\rangle$.
Lastly for (1) you may show that every element of the form on the right has to be contained in the generated subgroup on the left, which shows $\supseteq$. On the other hand, if you can show that the right hand side forms a group, you immediately get $\subseteq$, proving equality. For the latter you will use the commutativity of the $a_i$, so the abelianess of the generated subgroup immediately follows from the proof of its explicit form.