9
votes
1answer
317 views

Finitely generated group with $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups?

Let $X_G$ be the number of normal subgroups of a group $G$. Are there examples of finitely generated groups $G$ where it is consistent to have $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups? Also ...
1
vote
0answers
125 views

Is a certain group, derivable from the surreal numbers, isomorphic to the surreal numbers?

Let's treat $\mathbf{No}$ as a group under addition, and forget its field structure for a little bit. I will define a "maximally Archimedean subgroup" of $\mathbf{No}$ as a subgroup which is ...
10
votes
3answers
2k views

A “mother of all groups”? What kind of structures have “mother of all”s?

For ordered fields, we have a “mother of all ordered fields”, the surreal numbers $\mathbf{No}$, a proper-class “field” which includes (an isomorphic copy of) every other ordered field as a subfield. ...
12
votes
1answer
386 views

Without choice, can every homomorphism from a profinite group to a finite group be continuous?

In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the ...
5
votes
1answer
210 views

Is the dual of the product of infinite cyclic groups a free abelian group ?

By a theorem of Specker, the group $Hom(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this ...
11
votes
1answer
402 views

counting non-isomorphic groups of a given cardinality

Given an infinite cardinal $\kappa,$ is there some nice way to construct $2^\kappa$ non-isomorphic groups of that cardinality? In the answer to this stackexchange question, there is a fairly ...
4
votes
1answer
553 views

Can every nonempty set carry abelian group structure? [duplicate]

Possible Duplicate: Does every non-empty set admit a group structure (in ZF)? Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group? If ...
3
votes
1answer
325 views

Universe-sized groups with only set-sized normal subgroups, their cardinality in a certain range

Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all ...
0
votes
0answers
189 views

How many minimal pair-wise coverings can there be?

Suppose I have a set of finite sets: $X = \{V_1, V_2, V_3\}$ where each set called $V_i$ in $X$ contains a number of symbols (i.e. $V_1 = \{a,b,c\}$). $Z$ contains all of the Cartesian products of ...
6
votes
2answers
401 views

Distinct, non-homeomorphic, profinite topologies on a given abstract group ?

Just a silly little question which arose in connection with infinite Galois groups and their Krull topology:- can a given abstract group be endowed with distinct, non-homeomorphic, profinite ...
15
votes
1answer
729 views

What is the largest Laver table which has been computed?

Richard Laver proved that there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a*1 \equiv a+1 \mod 2^n$$ $$a* (b* c) = (a* b) * (a * c).$$ This is the $n$th Laver table ...
1
vote
0answers
216 views

Modern books about orders and algebras on trees

Please help to find books about orders and algebras on trees. If there is no modern books, please advice good old ones! I'm more interested in finite trees (my current problem), but infinite ones are ...
4
votes
1answer
303 views

When are unions of isomorphic groups isomorphic?

I was thinking about how to prove $\operatorname{Br}(K)\cong H^2(\operatorname{Gal}(\bar{K}/K),\bar{K}^*)$ without having to introduce inductive limits and all the profinite stuff. So, I started ...
4
votes
2answers
328 views

Is every bounded representation of Z unitarisable when all sets are measurable?

For the purpuse of this question, a group is amenable iff there exists a Foelner sequence. Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded ...
5
votes
1answer
411 views

Is it possible to construct (without choice, even?) a non-finitely-generated group with no proper non-finitely-generated subgroup?

Is there a non-finitely-generated group each of whose proper subgroups is finitely generated? If so, what form of choice (if any) is required to construct such a group?
5
votes
3answers
488 views

covering groups by infinitely many cosets

The classical Neumann lemma states that if a group is covered by finitely many cosets, then at least one of these cosets is the coset of a subgroup of finite index. (Actually, the lemma says more, ...
10
votes
3answers
714 views

Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice.

It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the ...
6
votes
5answers
1k views

Naturally occuring groups with cardinality greater than the reals.

In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, ...
3
votes
1answer
643 views

Shelah's proof of the independence of the Whitehead Problem

In the beginning Shelah classifies all $\aleph_1$-free Abelian groups into 3 possibilities each of which is satisfied by some $\aleph_1$-free Abelian group and the classification depends on the group ...
7
votes
8answers
2k views

Is there a ground between Set Theory and Group Theory/Algebra?

It is well known that there are strong links between Set Theory and Topology/Real Analysis. For instance, the study of Suslin's Problem turns out to be a set theoretic problem, even though it started ...
38
votes
5answers
4k views

Which graphs are Cayley graphs?

Every group presentation determines the corresponding Cayley graph, which has a node for each group element, and arrows labeled with the generators to get from one group element to another. My main ...
52
votes
3answers
5k views

Does every non-empty set admit a group structure (in ZF)?

It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary ...
27
votes
4answers
2k views

Can the symmetric groups on sets of different cardinalities be isomorphic?

For any set X, let SX be the symmetric group on X, the group of permutations of X. My question is: Can there be two nonempty sets X and Y with different cardinalities, but for which SX is isomorphic ...
2
votes
2answers
369 views

Automorphisms of the totally ordered group Z^n with lexicographical order

It is easy to see that the totally ordered group Z (the integers) with the natural order has no non-trivial automorphisms. Is this also true for Z^n with the lexicographical order?