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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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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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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Derived series of a quotient with smaller derived length. [closed]

I understand that if I have a solvable group $G$, i.e. : $$G \trianglerighteq G^{(1)}\trianglerighteq \ldots \trianglerighteq G^{(n)}=\{e\}$$ (in this case the derived length is n), then the quotient ...
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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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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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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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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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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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Does there exist a non-trivial group that is both perfect and complete?

A group $G$ is called perfect iff $G’ = G$. A group $G$ is called complete iff $Z(G) = \{e\}$ and $Aut(G) \cong G$. Does there exist a non-trivial group $G$, that is both perfect and complete at the ...
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Any Subgroup containing commutator subgroup is normal.

I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as If $H$ is a subgroup containing commutator subgroup then $H$ is normal. ...
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Commutator subgroup $G'$ is a characteristic subgroup of $G$

For any group $G$, prove that the commutator subgroup $G'$ is a characteristic subgroup of $G$. Let $U=\{xyx^{-1}y^{-1}|x, y \in G\}$. Now $G'$ is the smallest subgroup of $G$ which contains $U$. We ...