Problem about the subgroup generated by the commutators.

Question

I want to make a question about this exercise.

Let $G$ be a group, and let $a$, $b$, be elements of $G$. We define the commutator of $a$ and $b$ as follows:

\begin{equation} [a,b]:=aba^{-1}b^{-1}. \end{equation}

Let $C:= \langle \{[a,b] \mid a,b\in G\} \rangle$ be the subgroup generated by all the commutators in $G$.

($i$) Prove that $\forall\ a,b,c\in G$, $[a,b]^{-1}=[b,a]\ $, and $c[a,b]c^{-1}=[cac^{-1},cbc^{-1}].$

($ii$) Using ($i$), prove that $C \lhd G$.

I proved ($i$), and I'm now stuck on $(ii)$. My try has been this:

We have to see that $\forall\ g\in C, [a,b]\in C$ it happens that \begin{equation} g^{-1}[a,b]g\in C. \end{equation}

But this is true, because by ($i$) we have that \begin{equation} g^{-1}[a,b]g= [g^{-1}ag,g^{-1}bg] \end{equation}

Now I would say that it's over, this element is always in $C$ and hence $C \lhd G$. Is this correct?

Thanks.

Answer 1

There's something missing from your argument. $C$ is normal if for all $c \in C$ and all $g \in G$ we have $g^{-1}cg \in C$. But $C$ is the subgroup generated by all commutators: it is (in general) not just the set of commutators. Elements of the subgroup generated by a set $S$ can be written as products $s_1,\ldots,s_n$ where each $s_i$ is either an element of $S$ or the inverse of an element of $S$.

This isn't a big problem for your argument, but you need something extra. Let $c$ be any element of $C$. Then $c$ equals a product $c_1 c_2 \cdots c_n$ where $c_i = [g_i,h_i]^{\epsilon_i}$ where $\epsilon_i= \pm 1$ Then $g^{-1}cg = \cdots$

Answer 2

Try to prove this first:

any element of $C$ is a finite product of integral powers of commutators of $G$

Thus it is enough to show for any element $g\in G$ $g(aba^{-1}b^{-1})g^{-1}\in C$

Now $g(aba^{-1}b^{-1})g^{-1}=((ga)b(ga)^{-1}b^{-1})bgb^{-1}g^{-1}\in C$