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The answer is Yes for the second question, about $(\exists x)(\forall y)w=1$. Following Christian Remling's idea: If a sentence like $$\exists x(\forall y)(yxy^{-1}x^2y^{-9}\dots=1)$$ holds in all fin …

This is an open problem. See Wikipedia article: http://en.m.wikipedia.org/wiki/Finite_lattice_representation_problem Palfy and Pudlak's result (see open-source description in Palfy's article Interva …

Suppose $x$ is a word over the alphabet $\{0,1\}$. Let $a$, $b$ be elements of the group Dih$_k$ for some $k$. Let $\varphi=\varphi_{a,b,k}$ be the map from words over $\{0,1\}$ to elements of the di …

Under the Generalized Continuum Hypothesis, $$2^{\aleph_\alpha}=\aleph_{\alpha+1}\quad(\forall\alpha),$$ sets with no structure (so automorphisms are just bijections) is an example. Namely, by Cardi …

Yes. It suffices to show that any free semigroup embeds in a group. For this I refer you to MO question 3235: Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, an …

This must be known or easy for some of you, but here goes: Suppose $f_0,f_1:[n]\to [n]$ are invertible functions, where $[n]=\{0,\dots,n-1\}$ is a set of $n$ elements. For a word $w=w_1\dots w_m\in\{ …

It is known that there are non-amenable groups not containing $F_2$, the free group on two generators. We can even have that every 2-generated subgroup is finite. But is there a non-amenable group $ …

Let $p_i$ denote the $i$th prime number, and let $\oplus$ be the recursive join on $\omega$. Let $\mathcal O$ be Kleene's $\Pi^1_1$-complete set and $\mathcal O'$ its Turing jump. For any $B$, let $G …

Let $n$ be an integer. Andreas Thom mentioned that Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ …

Just want to point out that an algorithm to answer, given $n$ and a presentation of a finite group $G$, whether $G$ is isomorphic to a finite subgroup of $SU(n)$, exists as a consequence of Tarski's T …