# Proofs on the Commutator subhgroups

I have the following information about the commutators:

Let $$G$$ be a group. An element $$x \in G$$ is called a commutator if $$x$$ can be written in the form $$aba^{−1}b^{−1}$$ for some $$a, b \in G$$. The subgroup of $$G$$ generated by all the commutators of $$G$$ is called the commutator (or derived) subgroup of $$G$$ and is denoted by $$[G, G]$$ or $$G′$$. I need to prove the following:

1. Show that $$G$$ is abelian if and only if $$G'=\{e\}.$$

2. Let $$G$$ be a group and let $$N$$ be a normal subgroup of $$G$$. Prove that $$G/N$$ is abelian if and only if $$G' \le N$$.

The first question I honestly have no idea what it's asking me to do. I know a commutator can be written as $$aba^{−1}b^{−1}$$ for some $$a, b \in G$$ and $$D_8 =\langle a,b\mid a^{4}=e=b^{2},bab=a^{3}\rangle$$, but I really don't know where to from here.

So I think I have the backwards direction but not the foward direction.

Let $$G'=\{e\}$$ and let $$aba^{-1}b^{-1}=e$$. We see $$(aba^{-1}b^{-1})b=eb$$ is $$aba^{-1}=b$$ and then $$(aba^{-1})a=(b)a$$ which is $$ab=ba$$ so $$G$$ is abelian.

The foward direction I am unsure of . . .

For this question I think I have the forward direction but not the backwards direction.

Let $$G/N$$ be abelian and let $$a,b$$ exist in $$G$$. We see $$(aN)(bN)(aN)^{-1}(bN)^{-1}=(aba^{-1}b^{-1})N=[a,b]N$$ so $$[a,b]$$ exists in $$N$$. Since $$N$$ has all commutators then we see $$G'\le N$$. Again the backwards direction I am unsure of...

• If $G$ is abelian, what is $aba^{-1}b^{-1}$? Oct 30 '19 at 18:36
• @ArturoMagidin then aba^{-1}b^{-1}=bab^{-1}a^{-1}? And yes it was a typo
– user710744
Oct 30 '19 at 18:41
• If $G$ is abelian, then $aba^{-1}b^{-1} = a(ba^{-1})b$, and $ba^{-1}=a^{-1}b$, and then.... Oct 30 '19 at 18:43
• If you multiply $a$ and $b^{-1}$ to both sides you e=e correct?
– user710744
Oct 30 '19 at 18:52
• Please ask one question at a time. Oct 30 '19 at 20:04

1. If $$G$$ is abelian then $$aba^{-1}b^{-1} = aa^{-1}bb^{-1} = e$$.
2. Assume $$N$$ is a normal subgroup of $$G$$ and that $$G' \leq N$$. Consider the quotient map $$G \rightarrow G/N$$. Then the subgroup $$G'$$ of $$G$$ lies in the kernel, i.e., all commutators get sent to zero. In other words, $$(G/N)'$$ is trivial. Thus, by 1., $$G/N$$ is abelian.