In my abstarct algebra class there is a two part question regarding $D_3$. The first part asks us to find a commutator subgroup $[D_3, D_3]$ of $D_3$.
I am unsure on where to begin but I will state what I know that can maybe help. I know $D_3$ $=$ $<r,f|r^3 = f^2 = e, fr = r^2f>.$ I also know that the commutator of $a$ and $b$ in $G$ is $[a,b] = a^{-1}b^{-1}ab$ and the commutator subgroup $[G,G]$ of $G$ is the set of all finite products of commutators of elements of $G$. The reflections and rotations are hard for me to understand, and any help is appreciated.
The second part asks us if $[D_3, D_3]$ is a normal subgroup of $D_3$. I know a subgroup $N$ of $G$ is a normal subgroup if $gN = Ng$ for all $g \in G$. Any help is appreciated, this is my first theoretical class so I am struggling.
Commutator of 2 elements [a,b]=$aba^{-1}b^{-1}$
Commutator of group is the subgroup generated by all possible commutators.
In case of Dihedral group , $D_n=<r,s>$ that is it is generated by rotation and reflection
If we have generating set of group then to find commutator of that group reduced to commutator generated by its generator
Here r and s are 2 generator
1) $[r,r]=e$
2)$[r,s]=rsr^{-1}s^{-1}=r^2$
3)$[s,s]=e$
SO $D_3'=<r^2>=${$e,r^2,r$}
Part 2: To show it is normal
You can do it now directly as you have explicit subgroup given above
But there is also another way to argue in this case
Lemma: any subgroup of index 2 is normal in that group
Here $|D_3:D_3'|=2$
Hence normal