If a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is Nilpotent

Group Theory Problems and Solutions

Problem 128

Let $G$ be a nilpotent group and let $H$ be a subgroup such that $H$ is a subgroup in the center $Z(G)$ of $G$.
Suppose that the quotient $G/H$ is nilpotent.

Then show that $G$ is also nilpotent.

 
LoadingAdd to solve later

Sponsored Links


Definition (Nilpotent Group)

We recall here the definition of a nilpotent group.
Let $G$ be group. Define $G^0=G$,
\[ G^1=[G, G]=\langle [x,y]:=xyx^{-1}y^{-1} \mid x, y \in G\rangle,\] and inductively define
\[G^i=[G^{i-1},G]=\langle [x,y] \mid x \in G^{i-1}, y \in G \rangle.\] Then we obtain so called the lower central series of $G$:
\[ G^{0} \triangleright G^{1} \triangleright \cdots \triangleright G^{i} \triangleright \cdots. \]

If there exists $m\in \Z$ such that $G^m=\{e\}$, then the group $G$ is called nilpotent.

Proof.

Consider the natural projection $p:G \to G/H$.
Then we have $p(G^i)=(G/H)^i$.

Since $G/H$ is nilpotent, there exists $m \in \Z$ such that $(G/H)^m=\{eH\}$.
Thus we obtain
\[p(G^m)=(G/H)^m=\{eH\}.\] Thus for any $g \in G^m$, $g \in H \subset Z(G)$.

It follows that for any $g \in G^m$, $x \in G$ we have $gxg^{-1}x^{-1}=e$.
Since the elements $gxg^{-1}x^{-1}$ are generators of $G^{m+1}=[G^m, G]$, we conclude that $G^{m+1}=\{e\}$ and $G$ is nilpotent.


LoadingAdd to solve later

Sponsored Links

More from my site

  • The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly BiggerThe Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$. Then prove that $H \subsetneq N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$.   Proof. Note that we always have $H \subset N_G(H)$. Hence our goal is to find an element in […]
  • Commutator Subgroup and Abelian Quotient GroupCommutator Subgroup and Abelian Quotient Group Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$. Let $N$ be a subgroup of $G$. Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.   Definitions. Recall that for any $a, b \in G$, the […]
  • A Simple Abelian Group if and only if the Order is a Prime NumberA Simple Abelian Group if and only if the Order is a Prime Number Let $G$ be a group. (Do not assume that $G$ is a finite group.) Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.   Definition. A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]
  • Two Quotients Groups are Abelian then Intersection Quotient is AbelianTwo Quotients Groups are Abelian then Intersection Quotient is Abelian Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups. Then show that the group \[G/(K \cap N)\] is also an abelian group.   Hint. We use the following fact to prove the problem. Lemma: For a […]
  • Normal Subgroups, Isomorphic Quotients, But Not IsomorphicNormal Subgroups, Isomorphic Quotients, But Not Isomorphic Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$. Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.   Proof. We give a […]
  • A Condition that a Commutator Group is a Normal SubgroupA Condition that a Commutator Group is a Normal Subgroup Let $H$ be a normal subgroup of a group $G$. Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$. Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$. In particular, the commutator subgroup $[G, G]$ is a normal subgroup of […]
  • Quotient Group of Abelian Group is AbelianQuotient Group of Abelian Group is Abelian Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$. Then prove that the quotient group $G/N$ is also an abelian group.   Proof. Each element of $G/N$ is a coset $aN$ for some $a\in G$. Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]
  • Infinite Cyclic Groups Do Not Have Composition SeriesInfinite Cyclic Groups Do Not Have Composition Series Let $G$ be an infinite cyclic group. Then show that $G$ does not have a composition series.   Proof. Let $G=\langle a \rangle$ and suppose that $G$ has a composition series \[G=G_0\rhd G_1 \rhd \cdots G_{m-1} \rhd G_m=\{e\},\] where $e$ is the identity element of […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Group Theory Problems and Solutions in Mathematics
Normal Subgroups, Isomorphic Quotients, But Not Isomorphic

Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$,...

Close