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In this preprint of William Carter from 2013, it appears that there are very few known classes of torsion-free groups which are not unique-product groups, and that for these groups the unit conjecture …

Re: "I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one." Let $G$ be the the integers, and $I$ a countable ind …

First, let's deal with the case when $(G,+)$ is finitely generated. By the fundamental theorem of finitely generated abelian groups, let's go ahead and assume that $G$ is given to us in the form $\bi …

I believe that your intended question was something along the lines of whether or not the standard axioms can be simplified or reduced. This is in the spirit of Exercise 13.4 in Jech's "Set Theory", …

I believe the answer is yes. Assume, by way of contradiction, that some finitely presented group cannot be so expressed. Then we can choose such a group $G$ where for any generating set of the form …

Let $F$ be a field, and let $f_1,f_2,\ldots, f_k\in R:=F\langle\langle x,y\rangle\rangle$ with $k\in \mathbb{N}$. Order monomials in $R$ by degree, and then lexicographically. Since the question con …

A group $G$ is called a u.p.-group (short for unique product group) if for all nonempty finite subsets $A,B\subseteq G$, there exists an element $g\in A \cdot B$ which can be uniquely written as a pro …

Over a field, the answer to the first question is no. An element of $A$ of order 4 in SL(2,R) must satisfy the equation $x^{4}-1$. But it also satisfies the characteristic polynomial $x^{2}-{\rm Tr} …

Suppose we have a finitely presented group $G$ with solvable word problem. (For instance, the command RWSGroup in Magma terminates giving us a finite [but possibly gigantic] rewrite system.) Is there …

(Just a long comment that doesn't directly answer the bold question, but does [I hope] address an implicit question.) What I've found fascinating in thinking about your answer from that other thread …

If we modify your first question only slightly, then the answer to the question is no, there is no algorithm. By Theorem 1 of the paper G. Baumslag, W. W. Boone and B. H. Neumann, Some unsolvable pro …

More generally, let $k$ be any field, let $K/k$ be any separable algebraic extension, and let $R$ be any $k$-algebra. Then $J(K\otimes_k R)=K\otimes_k J(R)$. [See Theorem 5.17 in Lam's "First Course …

I've had occasions where I needed to know that some structure is "closed" under $S_n$. It is very convenient to only check that it is closed under those two, specific permutations. Afterwards, I can …

Let $f\colon S\to T$ be any function. There is the obvious refinement of $f$, by replacing the codomain $T$ with the image. Thus, every function factors into a surjection followed by an injection (a …

Let $S$ be a cancellative duo semigroup. Let $a,b\in S$ be arbitrary, and consider the element $x:=ba$. Clearly, $x\in Sa$ and by the duo property $x\in bS=Sb$. Thus, $S$ is right reversible, and s …