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Questions about the branch of algebra that deals with groups.
2
votes
1
answer
618
views
Cubical Complexes and Bass-Serre theory
I have been reading through Wise's lecture notes on cubical complexes, which summarises the proof of the virtual Haken conjecture and the proof that all one-relator groups with torsion are residually …
6
votes
0
answers
251
views
Recursive presentability of outer automorphism group
Is it possible for a finitely generated, recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,
$G$ finitely generated, recursively pre …
0
votes
3
answers
649
views
Center of a Symmetric Group on an Infinite Set
Let $X$ be a set, $G$ the group of bijection on $X$. Then it is well-known that if $|X|\geq 3$, $Z(G)$ is trivial. However, I cannot see a way of extending this proof to $X$ being an infinite set (oth …
3
votes
free group of finite rank can contain free groups of infinite rank as a subgroup
There is a very nice theorem which states:
Thm: If $F$ is a free group on $n$ generators and $H$ is a normal subgroup of index $j$ then if both $n$ and $j$ are finite $H$ is a free group on $j(n-1)+1 …
3
votes
Centralizers in amalgamated free products
I wanted to answer this question to give a solid journal citation (as requested in the comments to HJRW's answer), but also I feel there is some historical interest here.
It seems that the result you …
16
votes
Accepted
Exponent of a group
The group defined by $\langle x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$ has order 27, exponent 3 and is non-abelian.
(Checking exponent 3 basically comes down to ensuring that $( …
4
votes
Accepted
infinite groups in which all proper subgroups are almost-malnormal
Clearly if $G$ is an infinite group such that every proper subgroup of $G$ is finite then every subgroup of $G$ is almost malnormal.
Tarski monster groups are examples of such groups: Tarski monster …
5
votes
1
answer
261
views
Finite generation of intersections of free subgroups
Let $G$ be a finitely generated group, and let $F_1, F_2$ be two subgroups of $G$ which are free of finite rank at least 2. I am wondering what conditions can be placed on $G$ so that $F_1\cap F_2$ is …
11
votes
3
answers
1k
views
Solutions to some equations in a free group
Let $F$ be the free group on (say) two generators, $a$ and $b$. Let $A$ and $B$ be (freely reduced) elements of $F$. Let $W(X, Y)$ denote a word on the words $X, Y$.
-Is it ever true that the equatio …
8
votes
3
answers
735
views
Torsion in triangle groups
A triangle group has a presentation of the form,
$G=\langle a, b; a^{\alpha}, b^{\beta}, c^{\gamma}, abc\rangle, \alpha, \beta, \gamma \geq 2$
(I believe that these are also called von Dyke groups, …
2
votes
1
answer
296
views
Decision problems and group representations
If one takes a group presentation then one can ask various questions of it, such as "is this element equal to the identity", "are these elements conjugate" etc. I was wondering if the solution to such …
4
votes
2
answers
337
views
When are isomorphic copies of the base group in an $HNN$-extension subgroups of the base gro...
Let $G=\langle H, t; A^t=B\rangle$ by an $HNN$-extension of $H$, $A$ and $B$ isomorpic subgroups of $H$ where conjugation by $t$ induces the isomorphism.
Assuming $H$ is a finite group it is a well-k …
17
votes
1
answer
751
views
Are groups with every proper, non-trivial subgroup infinite cyclic simple?
In the 1970s Ol'shanskii constructed a non-cyclic finitely generated group $G$ with the following properties:
Every proper, non-trivial subgroup of $G$ is infinite cyclic.
If $X^m=Y^n$ for $X, Y\in G …
10
votes
1
answer
684
views
Nielsen equivalence in one-relator groups
Let $X=(x_1, \ldots, x_n)$ be an $n$-tuple of elements of a given group $G$. Then two $n$-tuples $X$ and $Y$ are Nielsen equivalent if there exists an automorphism of the free group on $n$-generators, …
8
votes
1
answer
361
views
Linking the residual finiteness of $G$ with $Aut(G)$ or $Out(G)$
There is a classic result of Baumslag which states,
Thm: If $G$ is residually finite then so is $\operatorname{Aut}(G)$.
While Grossman proved the (essentially) analogous result for $\operatorname{O …