Is the dihedral group $D_n$ nilpotent? solvable? I'm trying to solve this problem but I've been trying to apply a couple of theorems but have been unsuccessful so far. Can anyone help me?
I read that a direct product of a finite number of nilpotent groups is nilpotent. Here the definition of a nilpotent group is one that has a central series. A comment in my book following this claim says If $G_{ij}$ is the $i^{th}$ term of a central…
Suppose $p$ is a prime number and $G$ is a finite group, such that $\Phi(G) = C_p \times C_p$ , where $\Phi$ denotes the Frattini subgroup. Is it always true, that $p^4$ divides $|G|$ ? This statement can be easily proved for $p$ -groups by seeing…
I know this should be obvious but somehow I can't seem to figure it out and it annoys me! My definition of nilpotent groups is the following: A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U<N_G(U)$ for…
If $G$ is an arbitrary group we denote its abelianization as $G^{ab} := \frac{G}{[G,G]}$ , where $[G,G]$ is the commutator. As an abelian group it is characterized by the following universal property: Let $\pi : G \to G^{ab}$ be the quotient…
Given: Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements. Show that $G$ is finite. Also would love to know if it's possible to show that an infinite finitely generated nilpotent group has an infinite…
Suppose $G$ is a nilpotent, finitely generated group such that it's abelianization has rank $r$ . How does one go on proving that $G$ has a finite-index subgroup $H$ with free abelianization of rank $r$ ?
Suppose that $G$ is a torsion-free nilpotent group. It is a classical fact that if $g,h \in G$ have powers that commute, then $g$ and $h$ must commute. So if there are integers $m,n$ such that $[g^m,h^n] = 1$ , then $[g,h] = 1$ . This is usually…
I am currently working with Mal'cev completions, using the following definition: Let $N$ be group that is Nilpotent Torsion-free Finitely generated Then the Mal'cev completion or radicable hull is the unique group $N^\mathbb{Q}$ (up to isomorphism)…
P. Hall proved that every finitely generated torsion-free nilpotent group can be faithfully represented by upper unitriangular matrices over $\mathbb Z$ . The most famous example is the integral Heisenberg group, which is free nilpotent, of rank $2$…
Let $G$ be a nilpotent group generated by $d$ elements. Is there a function $r(d)$ such that every (necessarily finitely generated) subgroup $H$ of $G$ can be generated by at most $r(d)$ elements? Thanks in advance!
So I thought the cleanest way to do this was to simply prove $G' = 1$ since if $G$ is cyclic $G' = 1$ and then $G \cong G/G'$ , but I got no where with this. My next idea was since $G$ is nilpotent I know it's the direct product of normal Sylow…
Let $G$ be a polycyclic group and assume that every finite quotient of $G$ is nilpotent. Then $G$ is nilpotent. First some preliminaries: Every infinite polycyclic group contains a free abelian normal subgroup. On a polycyclic group one can use…
Fitting's Lemma is: Let $M$ and $N$ be normal nilpotent subgroups of a group $G$. If $c$ and $d$ are the nilpotent classes of $M$ and $N$, the $L = MN$ is nilpotent of class at most $c+d$. There is an exercise on Derek J.S. Robinson's A Course in…
