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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Commutator Subgroup is a Subgroup of "$G^2$"

Show that the commutator subgroup $G'$ of a group $G$ is a subgroup of $G^2$ defined by $G^2=\{x^2:x \in G\}$. I've tried writing the general element of $G'$; $hgh^{-1}g^{-1}$ as the square of some ...
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$R$ points of derived subgroup of algebraic group

I'm reading through Milne's book on algebraic groups, and in corollary 6.19 he writes: Assume that $G$ is affine or smooth, then (c) for all $k-$algebras $R$, $(\mathcal{D}G)(R)$ consists of the ...
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The only proper normal subgroups of a nonabelian quasisimple group are subgroups of its centre.

This is part of a bunch of exercises set by my academic supervisor. As such, I'm not sure whether all the hypotheses are needed for the conclusion. Please note that hints are preferred. The Question: ...
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Prove/disprove: Let $G$ s.t $|G| = p^3 \implies |G'| \leq p$. [duplicate]

Let $G$ s.t $|G| = p^3$ for some prime $p$. Prove or disprove: $|G'| \leq p$. I couldn't think of any counter-examples so I started proving this and I'm stuck unfortunately and was hoping to seek ...
MathStudent101's user avatar
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Solvable Groups and Derived Subgroups. Why does $D(G_i) \subset G_{i + 1} \implies D_i(G) \subset G_i$.

I'm working through the proof of the following theorem: $G$ is a solvable group $\Leftrightarrow$ There exists $n$ such that $D_n(G) = {1}$, where $D_i(G)$ is the $i$th derived subgroup of $G$. Proof: ...
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$G$ finite metabelian group, $P$ a Sylow $p$-subgroup. Show that $P'$ is abelian and normal in $G$

Let $G$ be a finite metabelian group and let $P$ be a Sylow $p$-subgroup of $G$. I have to observe that the derived subgroup $P'$ is abelian and normal in $G$. Since $G'$ is abelian, the subgroup $P' \...
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When is the special linear group $SL_n(\Bbb F_q)$ over a finite field $\Bbb F_q$ solvable?

Recall that the special linear group $$SL_n(\Bbb F_q)=\{ A\in GL_n(\Bbb F_q)\mid \det(A)=1\},$$ where $\Bbb F_q$ is the field of $q$ elements for finite $q$. A group $G$ is solvable if the derived ...
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If $G^{(i)}$ denotes the ith derived subgroup of $G$ then $G^{(i+1)}$ is a proper subgroup of $G^{(i)}$

Is this statement true? If $G^{(i)}$ denotes the ith derived subgroup of $G$ then $G^{(i+1)}$ is a proper subgroup of $G^{(i)}$ for all $i$. I think this statement is false by the symmetric group. ...
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If $G/G’ $ is cyclic, prove by induction that $G/Z(G)$ is cyclic if $G$ is $p$ group. [duplicate]

Let $G$ be a $p$ group. If $G/G’$ is cyclic, prove $G/Z(G)$ is cyclic by induction. I tried playing with the base cases : if $|G|=p $ or $p^2$, $G $ is abelian, so $G/Z(G)=G/G =\{G\}$ which is ...
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Testing if an element is in the derived subgroup

Let $ G $ be a finite group. Let $ G':=[G,G] $ be the derived subgroup. How do you test if an element $ g \in G $ is in $ G' $? I'm specifically interested in how to do this in GAP. In other words I ...
Ian Gershon Teixeira's user avatar
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What is known about this generalisation of the derived group?

Recently I've been revising a course on Representation Theory. As part of that course, we ended up proving that the intersection of the kernels of all 1D representations of a group is its derived ...
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The commutator of Holomorph of generalized quaternion is abelian?

Let $Q_{2^{n}} = \langle x, y | x^{2^{n-1}}=y^4 = 1, x^{2^{n-2}}=y^2, y^{-1}xy = x^{-1} \rangle$ - generalized quaternion group of order $2^{n}$. $\operatorname{Hol}(Q_{2^{n}})$ - Holomorph of this ...
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If $F$ is free and $R$ is normal in $F$, then $F/R'$ is torsion-free, where $R'=[R,R]$

If $F$ is free and $R$ is normal in $F$, then $F/R'$ is torsion-free, where $R'=[R,R]$. This is Exercise 11.50 in Rotman's An Introduction to the Theory of Groups with the following hint attributed ...
I Eat Groups's user avatar
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In $G/K(G)$ for commutator group $K(G)$, it is said that $[a][b]=[a][b][b^{-1}a^{-1}ba]=[a][b][b^{-1}][a][b][a]=[b][a].$ Why the first equality?

Let $K(G)$ be the commutator subgroup. It is said that in the quotient space $G/K(G)$ $$\begin{align} [a][b]&=[a][b][b^{-1}a^{-1}ba]\\ &=[a][b][b^{-1}][a][b][a]\\ &=[b][a]. \end{align}$$ ...
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Let $H = \langle X \rangle$ then $H' = \langle [x_1, x_2]^h \mid x_1, x_2 \in X, h \in H \rangle$ [duplicate]

Let $H = \langle X \rangle$, then $$H' = \langle [x_1, x_2]^h \mid x_1, x_2 \in X, h \in H \rangle.$$ Obviously $$\langle [x_1, x_2]^h \mid x_1, x_2 \in X, h \in H \rangle \leq H'.$$ Any tips on how ...
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Commutator subgroup of connected group.

Let $G$ be a compact, connected (Lie) group. I read in some paper that the commutator subgroup, $[G,G]$, which is the subgroup of $G$ generated by all its commutators, is also connected. Elements of $[...
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Request vetting of understanding of problem on commutator subgroups.

Let there be dihedral group $D_n$ and take two elements with first ($a$) being a generator rotation and other ($b$) any reflection. Which group is $[D_n,D_n]$? Till now not covered quotient groups, ...
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$\{ a \mid a\not \in [G,G],\ a \in Z(G) \} \cup \{e\}$ is a group or not

Let $G$ be a group. I want to check whether the set $$\{ a \mid a\not \in [G,G],\ a \in Z(G) \} \cup \{e\}$$ is a subgroup or not, where $[G,G]$ and $Z(G)$ are the commutator subgroup and center of ...
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Quick question about metabelian group

The following classic exercise question: show that the group $G$ is metabelian iff $G'' = e$ ($G''$ denotes $(G')'$, the commutator subgroup of the commutator subgroup of $G$). I know that the ...
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A problem in group theory and representation theory.

Suppose that $H = \{1, h, h^2, \ldots, h^{n-1}\}$ is a normal subgroup of a finite non-abelian group $G$ having order $n$. It is known that $H$ is cyclic with generator $h$. Let $c_G$ be the number ...
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Show that the twisted commutator, $T_g = \{x \in G \mid \exists k\in\Bbb Z, xgx^{-1} = g^k\}\le G$ for some $g \in G$ of finite order

I'm studying for qualifying exams, and this problem has me stumped. It should be straightforward (the definitions are fundamental), but I can't figure out the trick. Let $G$ be any group and let $g$ ...
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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 ...
Mikel Solaguren's user avatar
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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 ...
Mikel Solaguren's user avatar
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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 $\...
Mikel Solaguren's user avatar
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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 ...
Algebro1000's user avatar
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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 ...
Gillyweeds's user avatar
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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 ...
math seeker's user avatar
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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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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 ...
not all wrong's user avatar
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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 ...
Alejandro Bergasa Alonso's user avatar
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1 answer
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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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2 votes
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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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