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Questions about the branch of algebra that deals with groups.
11
votes
Accepted
Does every cancellative duo semigroup embed into a group?
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 …
28
votes
2
answers
849
views
$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)?
Edited to add: As no answers are forthcoming, does anyone …
4
votes
Is orientability a miracle?
(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 …
15
votes
Conceptual reason why the sign of a permutation is well-defined?
There are already plenty of other great answers, but I figured I might as well give my own.
My approach is to introduce students to permutations by taking plastic cups that have large numbers on them, …
11
votes
What is the standard 2-generating set of the symmetric group good for?
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 …
8
votes
1
answer
1k
views
First isomorphism theorem for sets?
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 …
5
votes
Accepted
Integral monoid rings and Ore conditions
Let $s\in S$ and $r\in \mathbb{Z}[S]$. We can then write $r=\sum_{i=1}^{n}\alpha_i t_i$ for some $\alpha_{i}\in \mathbb{Z}$ and some $t_i\in S$.
The left Ore condition on $S$ implies that there exist …
10
votes
Accepted
Is every finitely-presentable group a finite colimit of copies of $F_2$?
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 …
4
votes
Accepted
In the von Neumann–Bernays–Gödel axiomatic system, the Axiom of Transposition can be simplified
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", …
5
votes
0
answers
625
views
Unique product groups (and semigroups)
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 …
4
votes
Finding a compatible multiplication for a given group
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 …
11
votes
2
answers
568
views
Identifying a group without 2-torsion
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 …
3
votes
Jacobson radical of group algebra
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 …
4
votes
Accepted
presentations of subalgebras
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 …
5
votes
Accepted
A Non-Commutative Nullstellensatz
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 …