Proofs on the Commutator subhgroups

abelian-groupsabstract-algebrafinite-groupsgroup-theory

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…

Best Answer

  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.
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