Understanding a lemma about a finitely generated group

Question

There is a "lemma" in the book about group theory I'm seeing that states the following:

A group $G$ is said to be finitely generated if $G$ contains a finite subset $S$ such that $G = \langle S \rangle$. Note that if $S:=\{s_1,...,s_k\}$ and $G = \langle S \rangle$ is abelian then $$G=\{s_1^{n_1}...s_k^{n_k}:n_j \in \mathbb{Z}\}$$

There is no proof about this, and I don't understand it. Any hints?

Answer 1

I see by a comment in another answer that you are using a slightly different definition of $<S>$ as I am.

I'm using if $<S> = \{\prod t_i\}$ where $t_i$ is either a $s\in S$ or a $s^{-1}$ of an $s\in S$ and the product can be any combination thereof.

If $G$ is abelian then if you "clump" the same $s_i$s and $s_i^{-1}$ to be sequential then $<S>= \{\prod s_k^{n_i} \}$ ($n\in \mathbb Z$).

But you are using $<S>$ is the smallest group that contains $S$.

If $G$ is abelian it's easy to see $\{\prod s_k^{n_k} \}$ is a group. ($e = s_1^0$ and for every $g=\prod s_k^{n_k}$ then $g^{-1} = \prod s_k^{-n_k}$)

And if for any supgroup $K$ if $S\subset K$ then every $\prod s_k^{n_k}\in K$. so $\{\prod s_k^{n_k} \}\subset K$.

So $\{\prod s_k^{n_k} \}$ is the smallest subgroup containing $S$.

====

Well if $S= \{s_1, ...., s_k\}$ then $<S>$ then $G$ consists $s_{k_1}^{a_1}*s_{k_2}^{a_2}*s_{k_3}^{a_3}*.......*s_{k_z}^{a_z}$ where $a_i = \pm 1$ $\{k_1,k_2,k_3\}$ can be any sequence of indexes including the possibility of $k_{n+1} = k_n$. These are any product of elements of $S$ and inverses of elements of $S$.

If $G$ is abelian then we can rearrange all the component terms so elements are next to each other.

So if $n =s_{k_1}^{a_1}*s_{k_2}^{a_2}*s_{k_3}^{a_3}*.......$ we can rearrange the terms to $n = \underbrace{s_1^{a_{1_1}}*s_1^{a_{1_2}}*.... s_1^{a_{1_{n_1}}}}\underbrace{s_2^{a_{2_1}}*s_2^{a_{2_2}}*.... s_2^{a_{2_{n_2}}}}.....\underbrace{s_k^{a_{k_1}}*s_k^{a_{k_2}}*.... s_m^{a_{k_{n_k}}}}=s_1^{n_1}*s_2^{n_2}*...*s_k^{n_k}$

Answer 2

Since $G=\langle S\rangle$ is abelian and $S\subseteq G$, we have $s_is_j=s_js_i$ for all $1\le i, j\le k$. Thus we can move each of the generators past one another, so that we are left with $$G=\{s_1^{n_1}...s_k^{n_k}:n_j \in \mathbb{Z}\}$$ because every element of $G$ can be written in terms of the generators.