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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.


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 $N \in \mathbb N$


Would this be correct?

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  • $\begingroup$ You should also cover the case where $g$ cannot be written as the product of $2^N$ elements from $F$, for example if $g=f_1f_2f_3^{-1}$. $\endgroup$ Commented Feb 9 at 8:23
  • $\begingroup$ @anonymous67 We'd need the identity to be in the collection of sets then? Have identity $e \in G$. I didn't know if was subtly bundled into the definition already (or if it's pre-required that some $f_i = e$) due to how inverses combine and create $e$ on every second step. Would the expression $(\ggg)^N \big( \{f_i \} \cup \{ f_i^{-1}\} \cup \{ e \} \big)$ cover your case? $\endgroup$
    – Nate
    Commented Feb 9 at 8:52
  • $\begingroup$ "The Wikipedia article doesn't define the criteria for a generating set for a group in terms of logical propositions" I don't understand this sentence. In what terms it is defined then? It uses words instead of symbols, but that's the entire difference. $\endgroup$
    – freakish
    Commented Feb 9 at 8:56
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    $\begingroup$ @Nate again: just because one article utilizes words instead of symbols more than the other, it doesn't mean anything. Seriously. The wiki article is completely fine. In fact, it is more precise than your idea here, because you implicitly treat associativity here. $\endgroup$
    – freakish
    Commented Feb 9 at 9:01
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    $\begingroup$ @Nate the very first statement in the wiki article is first order logic proposition. Again: it doesn't matter it is written in words, instead of symbols. That doesn't make it less formal, it only makes it less symbolic. $\endgroup$
    – freakish
    Commented Feb 9 at 9:04

3 Answers 3

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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 $N \in \mathbb N$.

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  • $\begingroup$ That one's really good and after thinking it through your outline seems correct. I think it might be possible to streamline it even further -- not rigorously thought through, but it we use infix notations for things such as $+(a, b) = (a + b)$. If we extend the idea to $*(A \times A)$ we appear to have have: Iteration 0: $A$ Iteration 1: $A * A$ Iteration 3: $(A * A) * A$ And so forth. We might be able to ditch the helper function $(\ggg)$ entirely and end up with some solution of form $\underset{k}{ *} A$, similar in form to of course $\sum_k A$ from middle school algebra. $\endgroup$
    – Nate
    Commented Feb 9 at 11:14
  • $\begingroup$ I'll sleep on it. If I can't think of a faster streamlining in the morning I'll mark you as the answer. Else I'll toss the indexed series for $*(A \times A) = A * A$ back at you once I've thought about it more. $\endgroup$
    – Nate
    Commented Feb 9 at 11:20
  • $\begingroup$ You had a pretty good answer, but I after writing out your operation it seems we can "cut out" the middleman and just refer to the generating set without invoking $\ggg$. If I did this correctly then what I have is equivalent to your operation. $\endgroup$
    – Nate
    Commented Feb 9 at 21:49
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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 $G$.

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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 $N \in \mathbb N$.

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  • $\begingroup$ Note that $+$ is not necessarily commutative. I would use another symbol, but the math mode options on Stack Exchange are barebones, and I need $\times$ for the cartesian product. $\endgroup$
    – Nate
    Commented Feb 9 at 21:49
  • $\begingroup$ When talking about sets of this form, typically, we usually write just $S^n$ for a product of $S$ repeated $n$ times. (If there was a risk of confusion, I would write $S^{\cdot n}$, or $S^{+n}$ when writing additively.) Anyway, this definition is a bit cumbersome. The $\Leftarrow$ is unnecessary (since $S$ should be a subset of $G$, otherwise, the right hand side makes no sense). I think the statement on wiki ($G=\langle S\rangle$) is far more clear. $\endgroup$
    – tomasz
    Commented Feb 9 at 21:59
  • $\begingroup$ I think that $\Leftarrow$ holds though. If I take a generating set and apply the operation to it, any element created by that generating set would be in the original group $G$, yeah? $\endgroup$
    – Nate
    Commented Feb 9 at 22:01
  • $\begingroup$ That's the point. It holds trivially, so it does not make much sense as a part of a definition. $\endgroup$
    – tomasz
    Commented Feb 9 at 22:02

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