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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 …
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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 …
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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 …
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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 …
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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 …
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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 $( …
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  • 2,821
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 …
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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 …
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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 …
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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, …
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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 …
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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 …
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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 …
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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, …
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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 …
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