# Property of abelianization

This relates to this old MO question which had not been addressed effectively, though I do not feel I phrased the inquiry in the most effective means (or uploaded it on the appropriate website)

Define the abelianization of a team $G$ to be the quotient team $G_{ab} := G/[G,G]$, where $[G,G]$ is the commutator subgroup. I need to know just how this definition indicates the adhering to building of abelianization.

Allow $\phi: G \to G_{ab}$ be the approved surjection. For any kind of abelian team $H$ and also homomorphism $f:G\to H$, there exists an one-of-a-kind homomorphism $F: G_{ab} \to H$ such that $f = F\circ \phi$.

This is the capacity to 'come down to a homomorphism' that I clarified so severely in my first blog post, however at the time it was the only terms I listened to being made use of.

I realize that these 2 interpretations are equal yet I have actually not yet seen an evidence, neither took care of to confirm it myself. Please do aim me to an on-line evidence if you recognize of one.

12
2022-06-07 14:39:23
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If $f : G \to H$ is a homomorphism to an abelian team, after that $f(ab) = f(a) f(b) = f(b) f(a) = f(ba)$, therefore $[a, b] \in \ker f$, therefore $[G, G] \subseteq \ker f$. Is the remainder clear from below?

12
2022-06-07 15:07:58
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This generalises to any kind of variety of various other scenarios: as an example, if $G$ is a team and also $n\gt 0$ a favorable integer, allow $N$ be the subgroup of $G$ created by all components of the kind $g^n$, with $g\in G$. After that $G/N$ has actually exponent $n$, and also if $\pi\colon G\to G/N$ is the approved surjection, after that for any kind of team $H$ of exponent $n$ and also any kind of team homomorphism $f\colon G\to H$, there exists an one-of-a-kind homomoprhism $F\colon G/N \to H$ such that $f=F\circ \pi$.

For the basic context:

Definition. Allow $F_{\infty}$ be the free team of countably boundless ranking, and also allow $w\in F_{\infty}$. We claim that $w$ is an identification of $G$ if and also just if for every single team homomorphism $f\colon F_{\infty}\to G$, we have $f(w)=1$. Equivalently, every analysis of $w$ in $G$ is the identification.

Definition. Allow $S\subseteq F_{\infty}$. The selection of teams established by $S$ is the collection of all teams $G$ for which $w$ is an identification of $G$ for every single $w\in S$.

Suggestion. Allow $S\subseteq F_{\infty}$. The selection of teams $\mathfrak{V}$ established by $S$ satisfies:

1. $\mathfrak{V}$ is shut under subgroups: if $G\in\mathfrak{V}$ and also $H\lt G$, after that $H\in\mathfrak{V}$
2. $\mathfrak{V}$ is shut under homomorphic photos: if $G\in\mathfrak{V}$ and also $\pi\colon G\to K$ is a surjective team homomorphism, after that $K\in\mathfrak{V}$
3. $\mathfrak{V}$ is shut under approximate straight items: if $\{G_i\}_{i\in I}$ is a family members of teams (of approximate dimension), and also $G_i\in\mathfrak{V}$ for every single $i\in I$, after that $\prod\limits_{i\in I}G_i\in\mathfrak{V}$

Birkhoff is HSP Theorem. Allow $\mathcal{C}$ be a nonempty class of teams. After that $\mathcal{C}$ is a selection of teams if and also just if $\mathcal{C}$ is shut under (H) homomorphic photos, (S) subgroups, and also (P) approximate straight items.

Suggestion. Allow $G$ be a team, and also allow $\mathfrak{V}$ be a selection of teams. After that there is a tiniest regular subgroup of $G$, $\mathfrak{V}(G)$, such that $G/\mathfrak{V}(G)\in \mathfrak{V}$. If $\mathfrak{V}$ is established as a selection by the set of identifications $S\subseteq F_{\infty}$, after that $\mathfrak{V}(G)$ is the subgroup of $G$ created by the photos of $S$ under all homomorphisms $F_{\infty}\to G$.

Evidence. Plainly, there goes to the very least one regular subgroup $N$ such that $G/N\in\mathfrak{V}$, particularly $N=G$. If $\{N_i\}$ is a family members of regular subgroups of $G$ such that $G/N_i\in\mathfrak{V}$ for every single $i$, after that $\prod (G/N_i)\in\mathfrak{V}$, given that it is an item of teams in $\mathfrak{V}$. And also the approved map of $G$ right into $\prod(G/N_i)$ has bit $\cap N_i$, therefore $G/N_i$ is isomorphic to a subgroup of a team in $\mathfrak{V}$, therefore hinges on $\mathfrak{V}$. Hence, we can take $\mathfrak{V}(G)$ to be the junction of all regular subgroups $N\triangleleft G$ such that $G/N\in\mathfrak{V}$.

For the 2nd summary, allow $f\colon F_{\infty}\to G$ be a homomorphism. After that $\pi\circ f\colon F_{\infty}\to G\to G/\mathfrak{V}(G)$ is a map from $F_{\infty}$ right into a team in $\mathfrak{V}$, and also given that $S$ establishes $\mathfrak{V}$, after that $S$ has to hinge on th ekernel of this map. Therefore $f(S)\subseteq \mathfrak{V}(G)$. That is, $\mathfrak{V}(G)$ has all photos of $S$ under homomorphisms $F_{\infty}\to G$. Currently allow $N$ be the subgroup created by all such photos. This team is regular, due to the fact that if $f\colon F_{\infty}\to G$ is any kind of homomorphism, after that $f(S)^g$ favors $S$ under the homomorphism $\varphi_{g}\circ f$ (where $\varphi_g$ is the internal automorphism of $G$ established by $g$), therefore $f(S)^g\subseteq N$ for all $g\in G$. This holds for all $f$, so the creating set of $N$ is mapped to itself by conjugation. This reveals that $N$ is regular ; and also $G/N\in\mathfrak{V}(G)$, due to the fact that every homomoprhism $F_{\infty}\to G/N$ raises to a homomorphism $F_{\infty}\to G$, and afterwards the photo of $S$ in $G/N$ is unimportant. Consequently, $\mathfrak{V}(G)\subseteq N$, confirming equal rights. QED

The regular subgroup $\mathfrak{V}(G)$ is called the spoken subgroup of $G$ representing $\mathfrak{V}$.

Theory. Allow $G$ be a team, and also allow $\mathfrak{V}$ be a selection of teams. After that $\mathfrak{V}(G)$ is qualified by the adhering to global building: $\mathfrak{V}(G)\triangleleft G$, $G/\mathfrak{V}(G)\in\mathfrak{V}$, and also for every single $H\in\mathfrak{V}$ and also every team homomorphism $f\colon G\to H$, there is an one-of-a-kind team homomorphism $F\colon G/\mathfrak{V}(G)\to H$ such that $f=F\circ \pi$, with $\pi\colon G\to G/\mathfrak{V}(G)$ the approved estimate.

Evidence. First, we show $\mathfrak{V}(G)$ has this building. Allow $H\in\mathfrak{V}(G)$, and also allow $f\colon G\to H$ be any kind of team homomorphism. By the Isomorphism Theorem, $G/\mathrm{ker}(f)\cong f(G)\lt H$ ; given that $f(G)$ is a subgroup of a team in $\mathfrak{V}$, after that $f(G)\in\mathfrak{V}$, therefore $G/\mathrm{ker}(f)\in\mathfrak{V}$ (it is isomorphic to a team in $\mathfrak{V}$). Given that $\mathfrak{V}(G)$ is the tiniest regular subgroup of $G$ with a ratio in $\mathfrak{V}$, after that $\mathfrak{V}(G)\subseteq\mathrm{ker}(f)$, by the Universal building of the ratio, $f$ variables distinctly via $G/\mathfrak{V}(G)$, generating $F$. That $\mathfrak{V}(G)\triangleleft G$ and also $G/\mathfrak{V}(G)\in\mathfrak{V}$ has actually currently been developed.

Ultimately, we show that a regular subgroup $N$ pleasing this global building remains in reality $\mathfrak{V}(G)$. We understand that $\mathfrak{V}(G)\subseteq N$ by building and construction. And also by the global building, the approved estimate $G\to G/\mathfrak{V}(G)$ variables via $G/N$, so $N\subseteq \mathfrak{V}(G)$, confirming equal rights. QED

Now, the class of all abelian team is a selection (shut under homomorphism photos, subgroups, and also approximate straight items). Actually, the class is established by the solitary identification $x^{-1}y^{-1}xy$: a team $G$ is abelian if and also just if for every single $g,h\in G$, $g^{-1}h^{-1}gh = 1$ ; if we allow $\mathfrak{A}$ represent the selection of all abelian groups, after that $\mathfrak{A}(G)$ is specifically the subgroup created by all values of $x^{-1}y^{-1}xy$, i.e., the commutator subgroup. So the commutator subgroup has actually the wanted global building, therefore $G^{\rm ab} = G/[G,G]$ has actually the wanted global building.

If you intend to change "abelian group" with "abelian team of exponent $n$", after that you would certainly make use of the set $S=\{x^{-1}y^{-1}xy, z^n\}$, and also make use of the subgroup of $G$ created by the commutators and also all $n$th powers. If you change "abelian group" with "nilpotent of class $c$", after that you change the commutator subgroup with the $(c+1)$st regard to the lower main collection, established by the word $$[[\cdots [x_1,x_2],x_3]\cdots x_{n+1}].$$ And so on.

See additionally this discussion on commutator-center duality for extra on selections and also spoken subgroups.

11
2022-06-07 15:06:14
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