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Suppose $G=\langle S\mathrel| R\rangle$ is a presentation of a group (or semigroup) such that:

  1. $G$ is not generated by any proper subset of $S$.
  2. For any $r\in R$ we have $\bigl\langle S\bigm| R\setminus\{r\}\bigr\rangle\neq G$.

Does such a presentation of $G$ have a name? (Note that I understand this presentation need not be unique, and maybe even need not exist, but it does exist for my $G$ and I don't want to reinvent the wheel. I also know that "minimal" is used when speaking about finitely generated groups, but mine is not one.)

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    The first one requirement might not even make sense, since $w$ might occur in some of the relations. – 2017-02-22
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    @TobiasKildetoft Ah right, I'll clarify this, thanks! – 2017-02-22
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    If $s$ does not occur in any element of $R$, then clearly removing it will result in a different group. – 2017-02-22
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    @TobiasKildetoft Yes. The question still makes sense (and obviously is more about the relations -- not quite a surprise). – 2017-02-22
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    The thing is that it does make sense to consider the subgroup generated by a subset of $S$, and we could remove an element if the resulting set still generates the whole group. We would just need to possibly change the relations in some way. – 2017-02-22
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    Ok, I see that I need to think about this for a minute, thanks again! – 2017-02-22
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    some groups, like $(\Bbb Q,+)$, don't even have minimal generating subsets. – 2017-02-22
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    I am not sure this is standard terminology, but I suspect a more fitting name would be an irredundant presentation - that is no $s\in S$ or $r\in R$ can be removed. Minimal suggest that $S$ and $R$ are of minimal order, which is not necessarily the case (e.g. $\mathbb{Z}$ has generating set $\{2,3\}$ which could be considered irredundant but not minimal). – 2017-02-22
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    Can you give a simple non-example? For example, is $\langle a, b; a^2, b^3, b^5\rangle$ a non-example? (The $b^5$ is redundant, but not immediately so.) How about $\langle a, b; a^2, b^3, b^4\rangle$? – 2017-02-22
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    @user1729 I don't see how $b^5$ is redundant, I just see that $b = (b^3)^2(b^5)^{-1} = 1$, so $G=\langle a | a^2\rangle$ is generated by $a$ and the original presentation violates the 1st condition. The same for the second group with $b = (b^3)^{-1} b^4 = 1$. – 2017-02-22
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    @yo' Sorry, typos! all the relations were meant to be powers of $b$. Then $b^2$ immediately implies $b^4$, while that $b^5$ follows from $b^2$ and $b^3$ is less immediate. – 2017-02-22
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    @user1729 But yes, these are exactly the cases I have in mind (both the original one where $b$ doesn't add anything and the corrected one where a relation can be derived from the other ones). – 2017-02-22

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I think the word you are looking for is aspherical.

There are (classically) 5 different definitions of an "aspherical presentation". They may all be stronger than what you need though; I give this as an answer because your presentation may be aspherical (and certainly being aspherical implies your property). The reference is: Chiswell, Collins and Huebschmann, Aspherical group presentations, Mathematische Zeitschrift 178, 1-36 (1981) MR0627092.

As a warning: this paper clarifies certain issues in one of the standard tests in geometric and combinatorial group theory (Chapter III of Lyndon and Schupp Combinatorial group theory). so do not look up that text!

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    Now, this is quite dense for my knowledge of group theory, but thanks, I'll try to get into this! – 2017-02-22
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    Any reason asphericity is relevant here? Consider the case of the presentation $$ for the infinite cyclic group. Or the presentation $$ for the trivial group. – 2017-02-22
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    @MoisheCohen I am not sure I understand your point. The presentation $\langle x, y; y=x^2\rangle$ is aspherical (torsion-free one-relator presentations are always aspherical; this is an old result of Lyndon). It also has the property we are talking about here. (Asphericity is certainly relevant, but may not be the correct answer.) – 2017-02-23
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    @user1729: Of course it is aspherical, but is also redundant in the sense of the question asked. Thus, asphericity does not imply "irredundant". Conversely, asphericity clearly does not imply "redundant". Then what is its relevance? – 2017-02-23