# Excercise with the normal subgroup $K$ of $G$ with $K:=\bigcap\{H \text{ subgroup of } G: \forall_{x,y\in G}:xyx^{-1}y^{-1}\in H\}$

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 $$K$$.

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

• I don't understand why $K\subseteq \ker(\phi)$ holds – RM777 Feb 2 '19 at 18:33
• 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. – drhab Feb 2 '19 at 18:37
• 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? – RM777 Feb 2 '19 at 18:41
• 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 $A=\cap\{H\mid S\subseteq H, H\text{ subgroup}\}\subseteq L$. – 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 $$\;\phi=\phi'\circ\pi\;$$