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This is a remark rather than an answer to your question. If you remove the word `simple' it is easy to find such pairs of finite groups. The first examples I learned were (I think) constructed by At …

For any finite presentation there is an algorithm that will tell you if a word is equal to the identity: 1. Fix some positive integer $n$. 2. Write down all words in the generators that can be writte …

Let $G_1$ be the wreath product of $\mathbb{Z}$ with $\mathbb{Z}$. There is a surjective homomorphism $G_1\rightarrow \mathbb{Z}$ whose kernel $G_2$ is isomorphic to $G_1$. This gives a sequence of …

Is there an algorithm to distinguish the elements of the set $S$? If so, here is a word problem algorithm. This doesn't seem to use any transitivity properties, just faithfulness. Start with the pos …

$\DeclareMathOperator\SO{SO}\SO(n,\mathbb{Q})$ is the group of $n\times n$ matrices $A$ with rational entries such that $AA^t=I$ and $\hbox{det}(A)=1$. The $n$ coordinate subgroups of $\SO(n,\mathbb{Q …

It is true that a group $G$ is $FP_2$ if and only if $G$ acts freely cellularly on a connected CW-complex with trivial first homology group. I see from the comments that you are worried about finite …

Tits' solution to the word problem for Coxeter groups involves making subsitutions $u\mapsto v$ where $|u|=|v|$ and $uv^{-1}\in R$. Many (most) Coxeter groups are not hyperbolic, so (by Andy Putman's …

This question for a product of one Gromov hyperbolic space is a famous open problem (see for example Bestvina's problem list) and as far as I am aware the general case is also open.

If you add the relations that the standard generators have order two, then instead of getting the braid group on $n$ strands, you get the symmetric group on $n$ symbols. The generator that correspond …

Let $G=(\mathbb{Q},+)$. Then ${\rm cd}(G)=2$ and ${\rm cd}(G\times G)=3$.

An alternative construction of such a group: let $G$ be the free product of two copies of $A_5$, i.e., $G=A_5*A_5$. This group $G$ is perfect and it is also virtually free, so that any torsion-free s …

The action of the quotient on the cohomology groups of the normal subgroup is the trivial action, because the normal subgroup is central. (Think of group cohomology as a functor of the group: the con …

No, not even in the case when $G$ has a 1-dimensional classifying space: if $G$ is free of rank at least two, then for any $N$ with $G/N\cong \mathbb{Z}$, $N$ will be free of infinite rank and so $N$ …

This answers the other part of your question, not answered by Thompson's group. For each $i\geq 3$ there is a finitely presented group $G_i$ with the property that $H_i(G_i\mathbb{Q})$ is infinite di …

Inspired by YCor's remark I realized that every finite abelian $p$-group $G$ of rank three will have essential elements in $H^3(G;U(1))$. I think that this works for $p=2$ as well, but here's an argu …