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Let there be dihedral group $D_n$ and take two elements with first ($a$) being a generator rotation and other ($b$) any reflection. Which group is $[D_n,D_n]$?

Till now not covered quotient groups, normal groups, or conjugates even.

Request vetting of understanding of the problem :

Problem:

I have some idea about commutator as:
Given $a,b\in G$ for a group $G$, define the commutator of group element/composition $ab$ to be $[a,b]=aba^{-1}b^{-1}= ab(ba)^{-1}$.

So, do need to check if each element $ab, (ba)^{-1}$ is a generator of the whole group $D_n$?

But, here states:

If $G$ is a group and $X, Y \subseteq G$ then the commutator subgroup of $G$ is defined as $[G, G] = \langle [x, y] \mid x, y\in G \rangle$, where $[x, y] = x^{-1}y^{-1}xy$ and the group generated by commutator elements from $X$ and $Y$ is $[X, Y] = \langle [x, y] \mid x\in X, y\in Y\rangle$.

By it, that means that:
Need to check if each element $a,b$ (in $ab$) is a generator of the whole group $D_n$?

The first interpretation seemed intuitive, as each of $ab, (ba)^{-1}$ is an element as well as an entry in the group table.

So, taking the two elements as seperately involved in the composition $ab$ didn't seem a natural way to interpret.

If am wrong (second interpretation is correct), then please tell.

jiten
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    Take all $x,y$ in $G$. Take $[x,y]$ for these, as a set. See what subgroup they generate. – David A. Craven Jul 13 '22 at 13:02
  • Let $D_2= {e, r, s, sr}$ be the set; the subgroups generated by all pairs of non-trivial elements are: $(s,r)= {r,s, sr,e}$, by $(sr,s)={sr,s,e}.$ – jiten Jul 13 '22 at 13:08
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    See this https://math.stackexchange.com/a/1065045/1070376. So $3)$. – suckling pig Jul 13 '22 at 13:10
  • @Cpc Thanks, but seems not enough to understand. If could see my last comment, please. – jiten Jul 13 '22 at 13:19
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    Go back to the start. What is $[r,s]$? – David A. Craven Jul 13 '22 at 13:23
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    Follow the link. It is proven that $D_n'=\langle r^2\rangle $. And when $n$ is odd, $\langle r^2\rangle =\langle r\rangle $. (Reason?) – suckling pig Jul 13 '22 at 13:25
  • @DavidA.Craven $[r,s]=rsr^{-1}s^{-1} = rsrs= e$ – jiten Jul 13 '22 at 13:26
  • It may help to know that if $r$ is any rotation, and $t$ is any reflection, then $rt=tr^{-1}$. – Gerry Myerson Jul 13 '22 at 13:31
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    "Till now not covered quotient groups, normal groups, or conjugates even". You seem to be very keen to learn the subject backwards. The whole point of the derived group is that it yields the largest abelian quotient - to start with the commutators and see what happens seems to me perverse. – ancient mathematician Jul 13 '22 at 13:36
  • Well, $rs$ is a reflection, so $(rs)^2=e$ is automatic. – Gerry Myerson Jul 13 '22 at 13:39
  • @ancientmathematician If course is like that, what can I say. But, why it seems perverse, if in the beginning this was introduced? – jiten Jul 13 '22 at 13:41
  • @DavidA.Craven Sorry, left one non-trivial subgroup formed by $(r,sr)= {r,e,sr,s}$. But, order matches to another by $(s,sr)$. – jiten Jul 13 '22 at 13:53
  • @Cpc It is possible that in context of this post, your answer might be better than other links. Please. – jiten Jul 13 '22 at 14:42
  • There's no need for a dupe answer here. The proof in the link is complete. Note, $n$ odd implies $(2,n)=1$, so $\langle r^2\rangle =\langle r\rangle $. – suckling pig Jul 13 '22 at 16:31
  • @Cpc You stated choice as : $3,$ though for $2\mid n$ have $\langle R^2\rangle = R^{2i}, 0\leq i \leq n/2.$. Isn't for $2\mid n$, need $\langle R\rangle$ to generate the cyclic group of rotations inside $D_n.$ Also, unless choice is $4$, then how are going to generate the reflection elements too, which are must to generate $D_n$? – jiten Jul 14 '22 at 02:01
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    You don't get $D_n$ for the commutator. $D_n$ isn't perfect. It doesn't equal its commutator. The commutator is $\langle R^2\rangle $. – suckling pig Jul 14 '22 at 02:21
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    $\langle R^2\rangle =\langle R\rangle $ for $n$ odd. For $n$ even you don't get the whole $\langle R\rangle $. – suckling pig Jul 14 '22 at 02:25
  • @Cpc Please suggest a source for quote that: $D_n$ is not perfect. But, coupling one rotation (non-trivial) with any reflection can generate all reflections. In choice 3, there is no reflection element in the commutator. – jiten Jul 14 '22 at 02:45
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    That $D_n'\not=D_n$ is clear from the proof in the link. There's no reflections, in particular. – suckling pig Jul 14 '22 at 02:52

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