So far I have showed that $K$ is a normal subgroup and that the Operation defined on $G/K$ is abelian.

Now I have to show that if $\phi:G\rightarrow A$ is a group homomorphism and $A$ is abelian. Then there exists a $\phi':G/K\rightarrow A$ such that $\phi$ can also be expressed as $\phi '\circ\pi$, where $\pi$ is the natural homomorphism $\pi:G\rightarrow G/K$.

I know since $\phi$ is a homomorphism that $G/\ker(\phi)\cong \text{Im}(\phi)$.

So there exists a bijection $f:G/\ker(\phi)\rightarrow \text{Im}(\phi)$.

But I don't know how to prove that $\ker(\phi)=K$.

What I do know is that

$$z\in\{xyx^{-1}y^{-1},x\wedge y\in G\}=:S\Longrightarrow z\in \ker(\phi)\Longrightarrow S\subseteq\ker(\phi)$$

Also by Definition of $K$, I know that $S\subseteq K$.

Now there might be elements in $K$ which cannot be expressed this way. Also there might be elements in the kernel which are not in $K$.

  • $\begingroup$ I suppose you meant to write "...such that $\;\phi\;$ can be expressed as $\;\color{red}{\phi'}\circ\pi\;$ ..." $\endgroup$ – DonAntonio Feb 2 '19 at 18:05
  • $\begingroup$ yes you are right $\endgroup$ – RM777 Feb 2 '19 at 18:08

If $K$ and $N$ are normal subgroups and $K\leq N$ then the map $G/K\to G/N$ prescribed by $gK\mapsto gN$ is well defined and is a group homomorphism.

Denoting this map by $\nu$ for the natural homomorphisms $\pi_N:G\to G/N$ and $\pi_K:G\to G/K$ we find:$$\pi_N=\nu\circ\pi_K\tag1$$

Now if $\phi:G\to A$ is a group homomorphism where $A$ is abelian then for the subgroup $K$ mentioned in your question we find: $K\leq N:=\mathsf{ker}\phi$.

As usual we can write $\phi=\psi\circ\pi_N$ for a group homomorphism $\psi:G/N\to A$ and applying $(1)$ we find:$$\phi=\psi\circ\nu\circ\pi_K=\phi'\circ\pi_K$$

Here $\phi'=\psi\circ\nu:G/K\to A$ is a composition of group homomorphisms, hence is a group homomorphism.

  • $\begingroup$ I don't understand why $K\subseteq \ker(\phi)$ holds $\endgroup$ – RM777 Feb 2 '19 at 18:33
  • $\begingroup$ It is a consequence of $S\subseteq\mathsf{ker}(\phi)$ (mentioned in your question as something you know). Note that $K$ is the smallest subgroup that contains $S$ as a subset. $\endgroup$ – drhab Feb 2 '19 at 18:37
  • $\begingroup$ So there is a General Statement if $A$ is defined as the smallest subgroup that contains a set $S$ then every subgroup which contains $S$ must also contain $A$. Could you prove this Statement really quick or give me a source where I can see this Lemma? $\endgroup$ – RM777 Feb 2 '19 at 18:41
  • 1
    $\begingroup$ In that situation: $A=\cap\{H\mid S\subseteq H, H\text{ subgroup}\}$. If $L$ is a subgroup that contains $S$ as a subset then $L\in\{H\mid S\subseteq H, H\text{ subgroup}\}$ and consequently $A=\cap\{H\mid S\subseteq H, H\text{ subgroup}\}\subseteq L$. $\endgroup$ – drhab Feb 2 '19 at 18:44

Observe that we know (or should know) that $\;G/N\;$ abelian iff $\;G'=[G,G]\le N\;$ , for $\;N\lhd G\;$. so

$$G/\ker\phi\cong\phi(G)\le A\;\text{(abelian)}\iff G'\le\ker\phi$$

But certainly $\;K\le G'\;$ as $\;G'\;$ contains all the products $\;[x,y]:=x^{1}y^{-1}xy\;$ (and all the elements generated by this kind of products), i.e.: $\;G'\;$ is one of the subgroups $\;H\;$ that appear in the intersections that is the definition of $\;K\;$, and we thus get that $\;K\le G'\le\ker \phi\;$ , from where we get that if we define

$$\phi':G/K\to A\;,\;\;\phi'(gK):=\phi g$$

Observe that

$$gK=xK\iff x^{-1}g\in K\le\ker\phi\implies\phi\left(x^{-1}g\right)=1\iff\phi'(xK)=\phi x=\phi g=\phi'(gK)$$

and $\;\phi'\;$ is well defined.

Now check that we indeed have $\;\phi=\phi'\circ\pi\;$


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