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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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Derived series of a square-free order group stabilizes

I have been studying from these notes on groups, rings and fields by Lenstra and I find myself struggling with problem 1.20 which states the following Let $G$ be a finite group of squarefree order. ...
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Commutators in $SL(2,\mathbb{Z})$

Let $SL'(2,\mathbb{Z})$ be the commutator subgroup of $SL(2, \mathbb{Z})$, that is, $SL'(2,\mathbb{Z})$ is the group generated by elements of the form $$ \{ ABA^{-1}B^{-1} : A, B \in \text{SL}(2, \...
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Commutator Subgroup of Thompson's Group F [closed]

I am looking for a proof that the commutator subgroup of F is simple, I have found lots of articles about the commutator subgroup in general but havent found anything directly involving F and am ...
realmada's user avatar
5 votes
1 answer
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When is the center of group contained in the derived subgroup

Let $N$ be a group. Assume that $N$ is torsion-free, finitely generated and nilpotent. I read somewhere that $$ Z(N) \subset [N,N] \iff N \text{ cannot be written as a direct product of groups } N = A ...
noparadise's user avatar
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1 answer
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Does there exist an element of order $4$ in $GL_2(\mathbb{Z})/GL_2(\mathbb{Z})'$?

For a group $G$, let $G'$ denote the commutator of the group $G$, and if $H \leq G$ the left cosets will be denoted as $gH$. Now, I understand the fact that $[SL_2(\mathbb{Z}):GL_2(\mathbb{Z})'] = 2.$ ...
Zen's user avatar
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Proving the isomorphism type of the commutator subgroup of the dihedral group $D_n$

For any $n$, to what group is the commutator subgroup of the dihedral group $D_n$ isomorphic to? My solution is below. I request verification, feedback, and improvements. In particular, can you help ...
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proving a commutator identity

I need to understand why the following identity is true: $$[XK, XH]= [X, X][X, K][X, H][K, H]$$ where $H,K$ are normal subgroups of some given group, and $X$ a subgroup (not necessarily normal). And $[...
NotaChoice's user avatar
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What is the relationship between Inner Automorphism group and Commutator subgroup? Why are they both about commutativity?

Let $(G,\cdot)$ be a group. If $G$ is Abelian, the commutator subgroup $[G,G]=\{ghg^{-1}h^{-1}|g,h\in G\}$ is trivial; otherwise, the commutator subgroup is not trivial and there is the Abelianisation ...
Emancipatrix's user avatar
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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 ...
Bavanesh B S's user avatar
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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 ...
Et-'s user avatar
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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: ...
Shaun's user avatar
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3 votes
2 answers
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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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1 answer
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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: ...
Apollonius's user avatar
4 votes
2 answers
106 views

$G$ 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 ...
Shaun's user avatar
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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 ...
Kadmos's user avatar
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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
3 votes
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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 ...
bidermeyer's user avatar
3 votes
1 answer
65 views

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
1 vote
1 answer
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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 ...
mathemagiker's user avatar
4 votes
1 answer
213 views

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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4 votes
3 answers
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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 ...
Priya Pandey's user avatar
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2 answers
73 views

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 ...
Seth's user avatar
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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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1 vote
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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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1 answer
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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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3 answers
926 views

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
7 votes
2 answers
698 views

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
1 vote
0 answers
40 views

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 $a,b \in G$, $$\...
Mikel Solaguren's user avatar
3 votes
0 answers
50 views

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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1 vote
0 answers
156 views

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
1 vote
0 answers
644 views

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
1 vote
1 answer
113 views

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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-1 votes
1 answer
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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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0 votes
0 answers
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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
1 vote
1 answer
202 views

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 ...
Zed's user avatar
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2 votes
2 answers
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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 $g_2$ ...
User7238's user avatar
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6 votes
2 answers
331 views

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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1 answer
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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 $G′=H′\times A′$ ?
S.O's user avatar
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2 answers
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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 $R$ ...
Shaun's user avatar
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5 votes
1 answer
272 views

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
2 votes
1 answer
238 views

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: ...
JustWannaKnow's user avatar
2 votes
1 answer
715 views

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 $1$ ...
Anil Bagchi.'s user avatar
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2 votes
1 answer
33 views

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 $\boldsymbol{\mathcal{V}}$...
m1m111's user avatar
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