I have the following problem:
Let $G=\langle a_1,a_2,\ldots,a_n\rangle$ be the multiplicative group generated by $a_1,a_2,...,a_n$. Prove that if $a_ia_j=a_j a_i$ $\forall i,j\in\{1,2,\ldots,n\}$, then $G$ is an abelian group.
I don't understand what is means by "the multiplicative group generated by $a_1,a_2,\ldots,a_n$"? Are the elements in $G$ products of the $a_i$'s? If so, how can I write an arbitrary element of $G$?
Yes you're right, an element in $G$ would be a string of the form $$ a_{i_1}^{\pm 1}a_{i_2}^{\pm1}\cdots a_{i_k}^{\pm1}, $$ where the $i,j \in \{ 1,2,\ldots , n\}$. Then you know if $a_ia_j = a_ja_i$ you can put every element into the form $$ a_1^{k_1} \cdots a_n^{k_n} $$ where $k_i \in \mathbb{Z}$.
This is because if we have an element in the first form, we can switch the order of any two elements in the string. Doing this, we can put all the $a_1$'s at the front, then all the $a_2$'s at the left $\textit{etc}$ until we have the element in the desired form.
As an example consider $$ a_3a_2a_1 $$ then we know $a_na_1 = a_1 a_2$ so then $$ a_3a_2a_1 = a_3a_1a_2 $$ but then $a_3a_1 = a_1a_3$ so we have $$ a_3a_2a_1 = a_1a_3a_2 $$ and again $a_3a_2 = a_2a_3$, thus $$ a_3a_2a_1 = a_1a_2a_3. $$
If $G$ is the multiplicative group generated by $a_1,a_2,\cdots,a_n$, this means that that every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses. Formally if $x$ is an arbitrary element of $G$ then $x$ is of the following form: $$a_{i_1}^{k_1}a_{i_2}^{k_2}\cdots a_{i_n}^{k_n} $$ where $k_1,\cdots,k_n\in \mathbb{Z}$ (the definition of negative powers as powers of inverses), and $i_1,i_1\cdots,i_n\in\{1,\cdots,n\}$
