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Let $\Psi:G \to H$ be a homomorphism between a group $G$ and an Abelian group $H$,then show :

$$[G,G] \le \text{ker}(\Psi)$$ Where $[G,G]$ is the commutator subgroup of $G$.

I came up with the question: Does $$ [G,G] \trianglelefteq \text{ker}(\Psi)$$

hold ?

My work:

let $x,y \in [G,G]$ then $x=[a,b]$ and $y=[c,d]$ for some $a,b,c,d \in G$,then:

$$\Psi(xy^{-1})=\Psi(x)\Psi(y^{-1})$$

From the fact that $[h,g]^{-1}=[g,h]$ it follows that:

$$\Psi(xy^{-1})=\Psi([a,b])\Psi([d,c])=\Psi(a)\Psi(b)\Psi(a)^{-1}\Psi(b)^{-1}\Psi(d)\Psi(c)\Psi(d)^{-1}\Psi(c)^{-1}$$

Since $\forall g \in G:\Psi(g) \in \Psi(G)=\text{Im}(G) \subseteq H$ and $H$ is Abelian hence:

$$\Psi(xy^{-1})=e_H$$ I conclude that $xy^{-1} \in \text{ker}(G)$ but this does not imply $xy^{-1} \in [G,G]$ ,on the other hand $\text{ker}(G) \trianglelefteq G$ we see that $xy^{-1} \in G$,again this is not useful.

So is my proof wrong or it's not true to claim that $[G,G] \trianglelefteq \text{ker}(\Psi)$?

  • What is your work trying to show? To show that $[G,G]$ is in the kernel of $\Psi$, you can simply note that $\Psi([a,b]) = [\Psi(a),\Psi(b)] = 0$ since the target is abelian. – Alex Ortiz Nov 03 '20 at 15:54
  • Sorry, I thought, showing that it is normal had been done, e.g., here. – Dietrich Burde Nov 03 '20 at 16:23
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    A very easy consequence of normality is normality in any subgroup which contains the normal subgroup. That is, $N\trianglelefteq G$ and $N\le H\le G\implies N\trianglelefteq H$. –  Nov 03 '20 at 16:47
  • I was using the fact that $[G,G]\trianglelefteq G$, which is well-known. –  Nov 04 '20 at 01:37

2 Answers2

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By definition, $[G,G]$ is generated by elements of the form $[x,y]=xyx^{-1}y^{-1}$ (and not $xy^{-1}$). It suffices to show that every commutator is in $\ker \psi$ to show that $[G,G]\subset \ker \psi$. But

\begin{align} \psi(xyx^{-1}y^{-1}) &= \psi(x)\psi(y)\psi(x^{-1})\psi(y^{-1})\\ &= \psi(x)\psi(x)^{-1}\psi(y)\psi(y)^{-1}\\ &= 1 \end{align} Because $H$ is abelian.

The fact that $[G,G]$ is normal is because every conjugate of a commutator is a commutator : \begin{align} z [x,y]z^{-1} &= zxyx^{-1}y^{-1}z^{-1}\\ &=zxz^{-1}zyz^{-1}(zxz^{-1})^{-1}(zyz^{-1})^{-1}\\ &= [zxz^{-1},zyz^{-1}] \end{align} Thus, if $g = [x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n] \in [G,G]$, and if $h \in G$ : \begin{align} hgh^{-1} &= h\left([x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n]\right)h^{-1}\\ &= h[x_1,y_1]h^{-1}h[x_2,y_2]h^{-1}\cdots h[x_n,y_n]h^{-1}\\ &= [hx_1h^{-1},hy_1h^{-1}]\cdot[hx_2h^{-1},hy_2h^{-1}]\cdots[hx_nh^{-1},hy_nh^{-1}] \in [G,G] \end{align}

Didier
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  • @AlexOrtiz Now, it is – Didier Nov 03 '20 at 16:00
  • The second part actually doesn't imply that the commutator is normal without noting that conjugation by an element of the group defines a homomorphism. Because not every element of the commutator subgroup is a commutator, in general. –  Nov 03 '20 at 17:28
  • @ChrisCuster Not every element of $[G,G]$ is a commutator, but is a finite product of commutators. The result is thus straightforward. For the sake of completeness, I edited the answer. – Didier Nov 03 '20 at 18:09
  • Because $[G,G]$ is the subgroup generated by the commutators. An element is by definition a product of commutators, but does not have to be a commutator itself. For example, $1$ generates the full group $\mathbb{Z}$, but not every integer is the unit number $1$. – Didier Nov 03 '20 at 18:53
  • They look like... finite product of commutators. We cannot say much, it depends on the group $G$. And saying $[G,G] = \langle [a,b]~|~ a,b\in G \rangle$ is exactly the definition because $\langle S \rangle $ means "the subgroup of $G$ generated by the subset $S$". – Didier Nov 03 '20 at 19:12
  • Let us continue this discussion in chat. – Didier Nov 03 '20 at 19:38
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Yes because $\Phi(G)\le H$ must be abelian, as a subgroup of an abelian group. By the first isomorphism theorem $\Phi(G)\cong G/\rm{ker}\Phi$. And, finally, $G/H$ abelian if and only if $G'\subset H$.

Furthermore, $G'\trianglelefteq G\implies G'\trianglelefteq H$.

  • I might be misunderstanding your answer, but the question is not whether the image of $G$ under $\Psi$ is a normal subgroup, but whether the commutator subgroup is a normal subgroup of the kernel of $\Psi$. – Alex Ortiz Nov 03 '20 at 15:55
  • I added some details. –  Nov 03 '20 at 15:58
  • As far as I can tell, this still does not show that the commutator is a normal subgroup of the kernel, merely that it is a subgroup of the kernel. – Alex Ortiz Nov 03 '20 at 15:59
  • Right. You hadn't written normal at first. But if follows from the fact that $G'\trianglelefteq G$. –  Nov 03 '20 at 16:10
  • I did write normal at first, but besides that it was the question in the OP. – Alex Ortiz Nov 03 '20 at 16:15
  • Oh, I confused you with the OP. The OP wrote $G'\le H$, which merely means subgroup. –  Nov 03 '20 at 16:16