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In the general case, to determine whether some relations imply another given relation is an undecidable problem. If it were decidable, then we could decide whether a given finite presentation was pres …

Let me give a totally useless pure-existence answer, while we wait for someone to show up with a better answer. Namely, if it is true that those relations imply x6 = 1, then there definitely will be …

This recent MO question, answered now several times over, inquired whether an infinite group can contain every finite group as a subgroup. The answer is yes by a variety of means. So let us raise the …

For question 1, it must be true for huge numbers of cardinal pairs, for the simple reason that there are only continuum many first order theories in a countable language, but there are more than conti …

This is not an answer, but merely an observation that there can be no computable procedure to transform any given finite presentation into a presentation that is optimal in your sense. The reason is t …

I would expect that automorphism groups of natural structures would count as natural groups in your sense. But automorphism groups of uncountable structures often have size larger than the continuum. …

The statement as you have written it is not true for any uncountable cardinal $\kappa$. To see this, let $G$ be any group of size $\aleph_{\beta+\omega}$, where $\kappa=\aleph_\beta$. Every such group …

Here is an example showing that the particular subgroups you describe need not be maximal. Consider the case of $\Pi_{n\in\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$, with the subgroup $H$ consisting of the …

Regarding the question is your last paragraph, there is the following often-studied but not-quite-equivalent-to-your property: A Boolean algebra $\mathbb{B}$ is almost homogeneous if for every nonze …

If the axiom of choice holds, then this is an immediate consequence of the upward Lowenheim-Skolem theorem. Any first order theory in a finite language with an infinite model, such as the theory of th …

The answer is yes, as a consequence of my answer to your other question. Namely, in that answer, we have a finite group presentation $H$ and a word $h$ such that the question $h=1$ in $H$ is independe …

The answer is yes. This is just an instance of the general phenomenon that every non-computable decision problem is saturated with logical independence. (See this related MO answer.) Theorem. If $A$ i …

The answer is no. For a counterexample, let $G_i=\mathbb{Z}$ be the integers and let $H_i=\frac1i\mathbb{Z}$, for positive natural numbers $i$. The union $\bigcup_i G_i=\mathbb{Z}$, but $\bigcup_i H …

The answer is a simple trick. Essentially no group theory is involved. Suppose that we are given a group presentation with a set of generators, and relations R_0, R_1, etc. that have been given by a …

Update. Here is a more direct construction. (See edit history for previous version.) There is such a universal computable group as you request. Let $F$ be the free group on infinitely many generators …