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The group $G$ is not solvable, since its quotient $$ \tilde{G} := \langle x, y \ | \ x^7 = 1, y^2xy = x^4, y^{15} = 1\rangle $$ is a group of order $423360$ such that $\tilde{G}'' \cong {\rm PSL}(3, …

The smallest groups in which the centralizer of every element is non-abelian have order $32$. You can find them with GAP as follows: gap> IsExample := G -> ForAll(ConjugacyClasses(G), > …

Your representation $p$ is not faithful, since we have $$ (ABA^{-1}BA^{-1}BAB^{-1})^3 \ = \ 1. $$ In particular, this means that $$ (aba^{-1}ba^{-1}bab^{-1})^3 \ = \ \left(\begin{array}{rr}% - …

The smallest counterexamples have order $16$. Up to isomorphism, there are $14$ groups of order $16$; these fall into $9$ distinct equivalence classes w.r.t. order portrait. The $3$ equivalence classe …

Your group $G$ is not solvable since it has a quotient isomorphic to ${\rm S}_5$. You can see this with GAP as follows: gap> F := FreeGroup("a","b","c"); <free group on the generators [ a, b, c ]> ga …

The counterexample described by Derek Holt can easily be checked with GAP as follows: gap> G := Image(IsomorphismPermGroup(PerfectGroup(960,2))); A5 2^4' gap> CommutatorLength(G); # > 1 => there are …

Question: For which "interesting" classes of finitely generated groups is it known whether every infinite group in the class has an element of infinite order? Some examples: For finitely gen …

The answer to both questions is no: Counterexample to first assertion: $G = {\rm A}_5$, $H = \langle (1,2,3,4,5) \rangle$. Counterexample to second assertion: $G = \langle (1,2,3,4), (1,3), (5,6) \r …

I don't have the book at hand, but I think the usual meaning of $G = A.B$ for groups $A$ and $B$ is that $G$ has a normal subgroup isomorphic to $A$ such that the quotient $G/A$ is isomorphic to $B$. …

There are no such groups. With GAP one can check this as follows: Construct the group $G := {\rm AGL}(4,3)$: gap> G := SemidirectProduct(GL(4,3),GF(3)^4); <matrix group of size 1965150720 with 3 gen …

You can obtain an embedding of $H$ into ${\rm GL}(2^r,\mathbb{Z})$ as follows: Given a positive integer $m$, put $$ A_m \ := \ \left( \begin{array}{ll} 0 & 1_m \\\ 1_m & 0 \ …

If you allow the group to be finite, any non-alternating finite simple group is a counterexample. Otherwise you can still obtain counterexamples from wreath products of such groups with the infinite c …

Meanwhile there is a paper which specifically refers to this MathOverflow question, and which in particular shows that in general, limit groups are not freely subgroup separable: Simon Heil, JSJ decom …

Let $G$ be a group, and assume that there exist $a, b, c \in G$ such that $abc$, $acb$, $bac$, $bca$, $cab$ and $cba$ are precisely 5 distinct elements (i.e. that precisely two of the products are equ …

What can be said about pairs of non-isomorphic groups which are epimorphic images of one another and which also embed into one another? Can such pairs of groups be 'classified' in some sufficiently w …