Questions tagged [derived-subgroup]

Also called the commutator subgroup of a group is the subgroup generated by all commutator elements of that group. Should be used with the (group-theory) tag.

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Closure of the commutator group of Lie solvable group

Help me please to understand the following. Let $G$ be a connected Lie solvable group. Let $G':=\overline{[G,G]}$ be a closer of the commutator group $[G,G]$. Is it possible that $G'=G$? If so what if ...
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Can we write a subgroup as a product of subgroups? [duplicate]

Let $G$ be a group such that $$G=\prod G_i,$$ where above product is arbitrary. Suppose that there exists a subgroup $H$ of $G$ such that $$\prod G_i=H\prod G_i',$$ where $G_i'$ is derived subgroup of ...
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Is there an easy way to find the derived group of $S_5$?

In order to find the derived group of $S_5$ I've tried using Lagrange’s Theorem to find the order of the possible subgroups but $O(S_5)=2^3\cdot 3 \cdot 5$ so there are too many possible subgroups to ...
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How to find the derived subgroup of $Q_8$

Given the group $Q_8=\{ \pm1, \pm i, \pm j, \pm k \}$ is there an easy way to find its derived subgroup? I've tried by hand and this seems such a big task to do manually. After this, by thinking a ...
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How to find the derived subgroup of a given group

Given a group $(G, \cdot)$ is there a way to find its derived subgroup other than calculating it by hand element by element? For instance, if a group is abelian you know that its derived subgroup is $\...
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Given $H \lhd G$ and $G/H$ is abelian then the derived group $G'$ or $[G,G]$ satisfies $G'\subseteq H$

Given $H \lhd G$ and $G/H$ is abelian then the derived group $G'$ or $[G,G]$ satisfies $G'\subseteq H$ I've approached this problem the following way: As $H$ is abelian then given $a,b \in G$, $$\...
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Intersection of the derived subgroup with a cyclic group

Let $G$ be a group and $G'$ be the derived subgroup. Let $C=\langle c \rangle $ be a cyclic subgroup of $G$. Is it true that the intersection $H=G'\cap C$ is trivial? I was thinking that if $g\in H$, ...
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Let $G$ be a group, $H\unlhd G$, prove that the commutator subgroup $H'$ of $H$ is a normal subgroup in $G$.

Let $G$ be a group, $H$, a normal subgroup of $G$, prove that the commutator subgroup $H'$ of $H$ is a normal subgroup in $G$. My ideas: I want to prove it like this: $gH'g^{-1} = H'$ $\forall g \in ...
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If every finitely generated subgroup of a linear group G is solvable, then G is solvable

I've been working on this question for a long time now and I still have no idea how to approach it. I tried induction on the dimension of the matrices in G but can't complete the induction step. Any ...
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The relationship between the commutator subgroup and the center of a group

I just started reading about the upper and lower central series of a group. I'm wondering if there is a general relationship between the commutator subgroup $G'$ and the center $Z(G)$ of a group $G$. ...
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If $G$ nilpotent and $G/G'$ is cyclic then $G$ is cyclic? [duplicate]

If $G$ nilpotent and $G/G'$ is cyclic then $G$ is cyclic? It is very easy to see that this is true when $G$ is finite: If $G$ is finite nilpotent then all maximal subgroups are normal, so $G/N$ has to ...
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why "this means that $G'' \subset Z(G')$"?

I am trying to understand the proof of the following question: Show that if $G'/G^{''}$ and $G^{''}/G^{'''}$ are both cyclic then $G^{''} = G^{'''}.$[you may assume $G^{'''} = 1.$ Then $G/G^{''}$ acts ...
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Why $G'' \triangleleft G$?

Here is the question I want to solve: Show that if $G'/G''$ and $G''/G'''$ are both cyclic then $G'' = G'''$. [you may assume $G''' = 1$. Then $G/G''$ acts by conjugation on the cyclic group $G''$. ...
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Information about $P'$ in $P \rtimes Q$

Let $p$ and $q$ be distinct primes. Suppose we have a group $G = P \rtimes Q,$ where $P$ is a non-abelian $p$-group and $Q$ is an abelian $q$-group such that it is a subgroup of ${\rm Aut}(P).$ Then ...
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Non-existence of perfect groups with order $180$

A group $G$ is perfect when $G=G'$. $\textbf{Question.}$ Prove that there are no perfect groups of order $180$. $\textbf{My attempt.}$ I assumed there is such a $G$, so it is not solvable. I was ...
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Can this group be a derived subgroup of any subgroup? [duplicate]

The subgroup of $G$ generated by the set $\{aba^{-1} b^{-1} \mid a,b \in G\}$ is called the commutator subgroup of $G$ and denoted $G'.$ There is no group $G$ such that $G' = S_4$. I think that the ...
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Let $H\leq G$ s.t. whenever 2 elements of $G$ are conjugate then the conjugating element can be chosen from within $H$. Prove that $G'\subseteq H$.

Question: Let $H\leq G$ such that whenever two elements of $G$ are conjugate, then the conjugating element can be chosen from within $H$. Prove that $G'\subseteq H$. Solution: So, if $g_1$ and $g_2$ ...
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If $G$ is a nilpotent group and $H\leq G$ with $H[G,G]=G$ then $H=G$.

I am studying for a qualifying exam and this problem has been a white whale. Let $G$ be a nilpotent group with subgroup $H\leq G$. If $H[G,G]=G$, then $H=G$. I believe I should use the fact that if $...
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The commutator of the product of 2 groups

Let $G=H\times A$ be the internal product of $A$ and $H$, I proved that if "$A$ is abelian then $G′=H′$ " But if we were not given that $A$ is abelian is $G′=H′\times A′$ ?
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Show ${\rm SL}(n,q)$ is perfect for odd prime power $q>3$.

This is Exercise 3.2.10 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this search and Approach0, it is new to MSE. The Details: On page 8, ibid., Let $R$ ...
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Is it "obvious" that nested commutators generate the derived series?

The derived series of a group is constructed iteratively, taking repeated commutator subgroups. A commutator subgroup is famously not only the set of commutators but the group they generate. This ...
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Number of conjugacy classes of $G/[G,G]$

Let $G$ be a finite group. We can define a relation $R$ on $G$ by setting $aRb \iff b = g^{-1} a g$ for some $g \in G$. The conjugacy class of an element $a \in G$ is defined as its equivalence class: ...
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What is the number of degree $1$ representations of a finite group?

Let $G$ be a finite group and $G'$ be the derived subgroup of $G.$ Then I know that every degree $1$ representation of $G$ factors through $G/G'.$ How does it imply that the number of degree $1$ ...
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Why do certain relatively free groups have perfect commutator subgroups?

I am trying to understand an argument in a group theory, which is not my strong suit (I mostly work with semigroups and rarely delve into the actual structure of groups). Let $\boldsymbol{\mathcal{V}}$...
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If $G=⨝_{i\in I} H_i$ and $K\unlhd G$ is perfect, then $K=⨝_{i\in I}(H_i\cap K)$.

This is Exercise 5.11(b) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE. The Details: On page 72 . . . Definition: If $\mathcal{F}=...
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The commutator subgroup of a nonabelian simple group $G$ is $G$ itself [duplicate]

I'm studying elementary level of algebra and I'm trying to prove that the commutator subgroup of a nonabelian simple group is the original group itself. It is trivially false if the group is abelian, ...
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Proves about $L_n$ (lower central series of a group $G$)

I'm trying to solve this problem from my abstract algebra course: The lower central series of a group $G$ is defined by means of $L_0=G$, $L_{n+1}=[G,Ln]=\langle[g,h]:g\in G,h\in L_n\rangle$, for any ...
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Let $G$ be a group. Prove that $G'$ is central iff ${\rm Inn}(G)$ is abelian.

This is Exercise 4.11 of Roman's "Fundamentals of Group Theory: An Advanced Approach." According to this search and Approach0, it is new to MSE. The Details: On page 33 of Roman's book, ...
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Prove that $D'_n=\langle x^2\rangle$

I want to check if my solution to one problem from my group theory course is valid. The problem is: Given $D_n=\{x^iy^j:0\leq i<n,0\leq j<2\}$, prove that $D'_n=\langle x^2\rangle$. My attempt ...
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If $G$ is group with $|G|=p^3$, $p$ prime, then $G'=Z(G)$

I'm trying to solve this problem from my abstract algebra text book: Being $G$ a non-abelian group of order $p^3$, with $p$ prime. Prove that $G'=Z(G)$ In my notation $G'$ is the derived subgroup of ...
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Assume $G$ is a finite group such that every maximal subgroup of $G$ is normal in $G$ and for $H \leq G$ we have that $HG’=G$,then show that $H=G$.

Assume $G$ is a finite group such that every maximal subgroup of $G$ is normal in $G$ and for $H \leq G$ we have that $HG’=G$,then show that $H=G$. Where $G^\prime$ is the commutator subgroup (AKA ...
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Condition that shows when there are elements in $[G,G]$ which are not commutator.(Looking for a proof)

It's known that for a given group $G$ and the derived subset $[G,G]$ it's not true that every element of $[G,G]$ is a commutator. I've seen a condition that shows when there are elements in $[G,G]$ ...
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The larger the commutator subgroup is, the "less abelian" the group is

On the page of commutator subgroup Wikipedia says that "the larger the commutator subgroup is, the "less abelian" the group is." I know that for every group $G$ and $ N\...
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$G$ is Abelian if and only if $[e,e]$ is the only commutator of $G$

Wikipedia claims that for a given group $G$ with the identity element $e$ the commutator $[e,e]=e$ is the only commutator if and only if $G$ is Abelian. I know that for a given $N \trianglelefteq G$ ...
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Show that $\forall n \in \mathbb N^+:G^{(n)}\trianglelefteq G^{(n-1)}$

Given a group $G$,the derived subgroup of $G$ denoted by $[G,G]$ is defined as $[G,G]=\langle [a,b]:a,b \in G\rangle$ ,where $[a,b]$ is called the commutator of $a$ and $b$. One can define: $G^{(0)}=G$...
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If $G$ is finite of odd order, then the product of all its elements, in any order, is an element of the commutator subgroup $G'$. [duplicate]

This is Exercise 3.36 of Roman's "Fundamentals of Group Theory: an Advanced Approach". According to Approach0, it is new to MSE. The Details: Neither semidirect products nor presentations ...
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If $G$ is a finite group with $G'<G$, then $G$ has a normal subgroup of prime index.

This is Exercise 3.8 of Roman's "Fundamentals of Group Theory: An Advanced Approach." According to Approach0, it is new to MSE. The Details: Definition: The derived subgroup $G'$ of a group ...
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2 votes
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Show that $[G,G]$ is a normal subgroup of $G.$

Here is the question I want to answer: In a group $G,$ the commutator of $x,y \in G$ is $[x,y] = xyx^{-1}y^{-1}.$ Let $[G,G]$ be the subgroup generated by all commutators in $G,$ noting that if $G$ is ...
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Is it true that the derived series of $G/G^{(n-1)}$ is of length $n-1$, where $n$ is the length of the derived series of $G$? [duplicate]

Consider a group $G$ and its derived series $$G=G^{(0)}\trianglerighteq G^{(1)} \trianglerighteq \ldots \trianglerighteq G^{(n-1)} \trianglerighteq G^{(n)}=\{1\}\,.$$ Is the derived series of $G/G^{(n-...
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How do I find the derived subgroup for this particular group?

In describing the character table for groups in the Representation and Characters of Groups by James, I came upon this particular group $G=\langle a,b: a^{2n} = b^4 = 1, ba = a^{-1}b^{-1},b^{-1}a=a^{-...
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Index of commutator subgroup in the commutator group

I need to prove (or find a counter example) that if $G$ is solvable and $H \leq G$ is a subgroup of finite index, then the commutator subgroup $D(H)$ is also a subgroup of finite index in $D(G)$. This ...
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If $N$ is a normal subgroup of $G$, and $N \cap [G,G]=\{e\}$, then $N$ is contained in $Z(G)$.

I found an answer to this here: If the intersection of a normal subgroup and the derived group is $\{e\}$, show that $N$ is a subset of $Z(G)$.. However I don't really understand some of the answers ...
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$[G,G] \leq \langle g_1,...,g_{n-1}\rangle$ if and only if $\langle g_1,...,g_{n-1}\rangle$ is normal in $G$

Let $G$ be a group generated by: $\{g_1,...,g_n\}$, then $[G,G] \leq \langle g_1,...,g_{n-1}\rangle$ if and only if $\langle g_1,...,g_{n-1}\rangle$ is normal in $G$. I think I have to prove that $...
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1 vote
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About -derived subgroup- of a group $G$

This question is already solved, I am typing down the resolution from manual solutions. However there are some steps that I did not understand properly, could anyone help me? P.S. My doubts are in ${\...
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Construct normal series from derived series through generating sets

This is from section 2 of the paper "Quantum algorithms for solvable groups" by J. Watrous, where $G$ is a solvable group, and $G = G^{(0)} \rhd G^{(1)} \rhd \cdots \rhd G^{(n)}$ is the derived series ...
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4 votes
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Product of a finite family of derived groups

Let $k\in\mathbb{N}$ and $(G_i)_{1\leq i\leq k}$ a finite sequence of groups. I want to prove $$\prod_{i=1}^k[G_i,G_i]=\left[\prod_{i=1}^k G_i,\prod_{i=1}^kG_i\right].$$ For $k=2$, I am able to derive ...
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Is $S_6$ the derived subgroup of some group?

I know that if $H$ is a complete group (meaning that the homomorphism $H\to\text{Aut}(H)$ is an isomorphism) and if $H$ is not perfect (meaning that $H^\prime\lneq H$) then $H$ is not the derived ...
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2 votes
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Is the class of $\omega$-soluble groups a variety?

Let’s call a group $G$ $\omega$-soluble if $\bigcap_{i = 1}^{\infty} G^{(i)}$ is trivial. Here $\{G^{(i)}\}_{i = 1}^\infty$ stands for the derived series of the group. Note, that not all $\omega$-...
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13 votes
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Can a group have a cyclical derived series?

Given any group $G$, one can consider its derived series $$G = G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\dots$$ where $G^{(k)}$ is the commutator subgroup of $G^{(k-1)}$. A group is perfect if $G=G^{(1)}$ ...
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Does there exist a finite group that is both perfect and immaculate?

A group $G$ is called perfect iff $G’ = G$. A finite group $G$ is called immaculate iff its order is equal to the sum of orders of its proper normal subgroups. Does there exist a finite group $G$, ...
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