# 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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### $G$ 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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### 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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### 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 ...
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. ...
### 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 ...