Is this a correct definition for generating set of a group?

Question

The Wikipedia article doesn't define the criteria for a generating set for a group in terms of logical propositions, so here's my attempt at it. It seems from (Why do generating sets need not contain the inverses of their elements?) and the Wikipedia article that the generating set doesn't need its inverses, so I tried to take that into account for the definition.

One other thing: It's a bit slippery to define the construction method, so we'll make a custom function $(\ggg )$ that makes this definition easier to write formally.

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Have a group $\mathfrak G = \big( G , * \big)$. Where $\left(G \times G \xrightarrow{\quad * \quad} G \right)$. For convenience, define a function $\ggg$ such that $$\ggg(A) = *(A \times A) $$ where $A$ is any set we want.

A generating set of $\mathfrak G$ is a set $F = \{f_i \} $ such that $$g \in G \iff g \in (\ggg)^N \big(\{ f_i\} \cup \{ f_i^{-1}\} \big) $$

For some $N \in \mathbb N$

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Would this be correct?

Answer 1

Your construction is close to be correct.

Have a group $\mathfrak G = \big( G , * \big)$ where $\left(G \times G \xrightarrow{\quad * \quad} G \right)$. For convenience, define functions $\gg_N$ recurrently such that $$\gg_1(A) = A $$ $$\gg_{N+1}(A) = *(\gg_N(A)\times A)$$ where $A$ is any set we want.

A generating set of $\mathfrak G$ is a set $F = \{f_i \} $ such that $$g \in G \iff g \in \gg_N \big(\{ f_i\} \cup \{ f_i^{-1}\} \big) $$

For some $N \in \mathbb N$.

Answer 2

Have a group $\mathfrak G = \big( G , + \big)$. For any set $S$, write $+ \left( S \times S \right)$ as infix $S +S$. A generating set of $\mathfrak G$ is a set $F = \{f_i \} $ such that $$g \in G \iff g \in \; \underset {k = 1} \sum \limits^N \left( \{ f_i \} \cup \{ f_i^{-1}\}\right) $$

for $N \in \mathbb N$.

Answer 3

Here is a different (but equivalent) approach. Recall that the intersection of any collection of subgroups of a group is itself a subgroup. Let $G$ be a group and let $S$ be any subset of $G$. Then the subgroup generated by $S$ is simply the intersection of all subgroups of $G$ which contain $S$ as a subset. We say that $G$ is generated by $S$ if this subgroup is the whole group $G$.