These 2 group theory declarations are "the same"?

Allow $G$ be a team and also $G'$ its commutator subgroup. Allow $\pi: G\to G/G'$ be the all-natural estimate.

Declaration 1 : $G/G'$ is the biggest Abelian ratio of $G$ in the feeling that if $H\unlhd G$ and also $G/H$ is Abelian, after that $G'\le H$. Alternatively, if $G'\le H$, after that $H\unlhd G$ and also $G/H$ is Abelian.

Declaration 2 : If $\varphi:G\to A$ is any kind of homomorphism of $G$ right into an Abelian team $A$, after that $\varphi$ variables via $G'$ ; i.e., $G'\le \ker{\varphi}$ and also there is a homomorphism $\hat{\varphi}:G/G'\to A$ such that $\varphi(g) = (\hat{\varphi}\circ \pi)(g)$. (That is, we have an expensive commutative layout.)

This is from Dummit and also Foote, p. 169, Proposition 7.

The evidence of (1) is really uncomplicated. Nonetheless, the writers assert that (1) is a restatement of (2) in regards to homomorphisms. Can any person clarify this? Due to the fact that it is unclear to me. Additionally, if I intended to confirm (2) outright, what should the map $\hat{\varphi}$ be? My first idea was specifying it as $\hat{\varphi}(aG')= \varphi(a)$, yet I do not assume this jobs.

Many thanks!

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2019-12-02 02:53:29
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Answers: 2

Your suggestion for (2) specifically functions. Any kind of 2 reps $a$, $b$ with $aG' = bG'$ will certainly be connected by a component $g' \in G'$ with $ag' = b$. After that, due to the fact that $G'$ is a part of the bit of $\varphi$, $$\varphi(b) = \varphi(ag') = \varphi(a)\varphi(g') = \varphi(a).$$

This actually is a restatement of the first isomorphism theory for teams to a "first homomorphism theory for groups", in a manner of speaking. The isomorphism theory claims that map $\theta: G \to A$ with bit $K$ will certainly factor via the quotient map $G \to G/K$ - - - yet actually for any kind of subgroup $N$ of $K$ with $N$ regular in $G$ will certainly offer a map $G \to G / N$ by the very same dish defined over.

In your details instance, when the target is abelian, the bit always has the commutator subgroup $G'$ of $G$, therefore you make use of the "first homomorphism theorem" to get your map $\hat\varphi: G / G' \to A$.

That the first isomorphism theory can be damaged this way suggests that, whereas the first isomorphism theory alone connects surjective maps off $G$ to regular subgroups of $G$, the weakening claims that the incorporations $N_1 \subseteq N_2$ of regular subgroups of $G$ is additionally reviewed the degree of homomorphisms.

Hope this aids!

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2019-12-03 04:22:08
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By the Homomorphism Theorem, any kind of homomorphism $f\colon G\to K$ variables via $G/\mathrm{ker}f$, suggesting that there is a map $\hat{f}\colon G/\mathrm{ker}f \to K$ such that $ f = \hat{f}\pi$. The map is without a doubt $\hat{f}(g\,\mathrm{ker}f) = f(g)$. This relates to the details instance given up Statement 2.

Modify : In reality, the complete Statement 1 is not equal to Statement 2, in the feeling that if you change $G'$ with an approximate subgroup $M$ of $G$ in both declarations, after that Statement 1 identifies $G'$, yet Statement 2 does not. That is, if you have

  • Statement 1' : If $H\triangleleft G$ and also $G/H$ is abelian, after that $M\subseteq H$ ; and also if $M\subseteq H$, after that $H\triangleleft G$ and also $G/H$ is abelian.

  • Declaration 2' : If $\varphi\colon G\to A$ is any kind of homomorphism of $G$ right into an abelian team $A$, after that $\varphi$ variables via $M$ ; that is, $M\subseteq \ker\varphi$ and also there is a homomorphism $\hat{\varphi}\colon G/M\to A$ such that $\varphi(g) = \hat{\varphi}\circ\pi(g)$.

The only subgroup $M$ of $G$ that pleases Statement 1' is $M=G'$. Nonetheless, any kind of subgroup of $G'$ that is regular in $G$ will certainly please Statement 2'.

Actually, Statement 2 amounts the first condition of Statement 1, particularly that if $H\triangleleft G$ and also $G/H$ is abelian, after that $G'\subseteq H$, plus the implied assertion that $G'$ itself is regular in $G$.

Thinking the first condition of Statement 1 plus the reality that $G'\triangleleft G$, if $\varphi\colon G\to A$ is a homomorphism, after that by the Homomorphism Theorem, allowing $H=\ker\varphi$, after that $G/H$ is (isomorphic to) a subgroup of $A$, therefore abelian, so we have to have $G'\subseteq H = \mathrm{ker}\varphi$ ; this is Statement 2 (with the last condition of 2 offered by the homomorphism theory as above).

Thinking Statement 2, (which unconditionally insists that $G'$ is regular) intend that $H$ is a regular subgroup of $G$ such that $G/H$ is abelian. After that taking into consideration $\pi\colon G\to G/H$ and also using 2, you end that $G'\subseteq \mathrm{ker}\pi = H$. And also normality of $G'$ adheres to from the declaration of 2, which needs it.

That is, they aren't fairly equal, due to the fact that Statement 1 has an additional condition, particularly the "Conversely ..." condition, which is not an effect of thinking Statement 2. Yet the first component of Statement 1 (plus "$G'\triangleleft G$") is equal to Statement 2.

Included earlier : To see that both are not fairly equal as mentioned, allow me offer you an instance of a subgroup $M$ of $G$ that pleases Statement 2' yet not Statement 1' : take into consideration the instance of $G=S_4$ ; after that $G' = A_4$. Currently allow $M = \{ 1, (12)(34), (13)(24), (14)(23)\}$. After that $M\triangleleft G$, and also the declaration in 2 holds for $M$ : offered any kind of homomorphism $f\colon G\to A$ with $A$ abelian, the map $f$ variables via $G/M$ and also there exists a homomorphism $\hat{f}\colon G/M\to A$ such that $f=\hat{f}\pi$. Nonetheless, $M$ is not the commutator subgroup of $G$. What is missing out on in Statement 2 for it to be a real matching of Statement 1 is some declaration that represents the assertion that $G/G'$ is itself abelian, which is what adheres to from the "Conversely ..." condition in Statement 1. One means to do it is to merely mention that $G/G'$ is itself abelian. An additional is to take into consideration the junction of all bits of all homomorphisms right into abelian teams, and also claim that $G'$ has to amount to that junction.

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2019-12-03 04:21:13
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