Does $[G,G] \trianglelefteq \text{ker}(\Psi)$ hold?

Question

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 $G$.

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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 $[G,G] \trianglelefteq \text{ker}(\Psi)$?

Answer 1

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}

Answer 2

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