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.

Given a group $G$, it's derived subgroup, also often called it's commutator subgroup is the subgroup generated by all elements of the form $xyx^{-1}y^{-1}$ in the group. This commutator element $xyx^{-1}y^{-1}$ is often written more simply as $[x,y]$, and subgroup is denoted commonly as $[G,G]$ or as $G'$ or as $G^{(1)}$.

The quotient $G/G'$ is the largest abelian quotient of $G$. We can see this by noting that by quotienting by the element $xyx^{-1}y^{-1}$ we're imposing the relation $xy = yx$ on $G$.

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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. I.e. we have to show that $\forall g\in G$ such…
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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 $G=G^{(1)}$ and thus has constant derived series, and solvable…
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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 need to show that $T(G') \subset G'$ for all…
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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 subgroup of any group. In particular, if $n\geq3$ and…
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Commutator Subgroup and the Largest Abelian Quotient

Good day. I'm not too sure how I should go about asking my question, so if it is in anyway confusing, feel free to edit it. In my Group Theory course we defined and showed a few properties of the commutator subgroup $[G,G]$ of a group $G$. And then…
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Showing commutator subgroup is a subgroup

Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this. Let $[a,b],[c,d] \in G'$, Then we have:…
user637978
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Generators of the first commutator subgroup

I found this statement in some lecture notes, and I am having trouble proving it, so I just want to make sure that I understand the statement: Let $G$ be a group generated by a subset $S$. Then the first commutator subgroup of $G$ is generated by…
Ludolila
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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 $\{1_G \}$. This is one simple example but is…
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Proving a commutator identity for groups

From Wikipedia: If the derived subgroup is central, then $$(xy)^n = x^ny^n{\left[y,x\right]}^{n \choose 2}.$$ I was able to prove this for the case $n = 2$: \begin{align*} (ab)^{-2}a^2b^2[b,a]&= b^{-1}a^{-1}b^{-1}a^{-1}a^2b^2b^{-1}a^{-1}ba\\\ &=…
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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 $$G=G_0 \triangleright G_1 \triangleright \dots…
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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 \times B \text{ where }A \text{ is non-trivial…
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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 $[G,G]$ have the form $[g_1,g'_1] \ldots…
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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 raises the question of whether the second derived…
not all wrong
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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 same time? Motivation behind this question: Both…
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Commutator subgroup of a dihedral group.

I have a few questions concerning an example of the commutator subgroups in the dihedral group. This example is found on pg.171 of Abstract Algebra by Dummit and Foote. Let $D_{2n}=\langle r,s |r^n=s^2=1, s^{-1}rs=r^{-1}\rangle$. Since…
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