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Re the close vote: There are only 2 questions here. 1 main question. 1 side question.

I'm trying to understand what's assumed vs proven in dihedral group. In particular I'm trying to understand dihedral group without thinking of rotations, reflections, etc.

Question 1: Is this correct, in particular where the order is part of the to-be-proven instead of part of the assumptions?

The assumptions for dihedral group (of order $2n$ but denoted $D_n$ instead of $D_{2n}$)are:

Let $n \ge 3$. Define $D_n := \langle r,s \rangle$, where $order(r)=n$ and $order(s)=2$ and $sr=r^{-1}s$. In particular, $D_n$ is completely determined by the preceding conditions. (Here I just pretend $D_n$ is some free group or whatever generated by letters $r$ and $s$ that just so happen to satisfy the 2 order equations and $sr=r^{-1}s$. I try not to think about rotations or reflections.)

The to-be-proven for dihedral group is:

Show that $D_n$ is of order 2n with $$D_n = \{1_{D_n},r,r^2, ..., r^{n-1},$$ $$s, rs, r^2s, ..., r^{n-1}s\},$$

The problem I see if you try to prove $D_n$ is of order $2n$ instead of assume it:

  1. For finite subgroups $H$ and $K$ of group $G$, we have $cardinality(HK) = \frac{order(H)order(K)}{order(H \cap K)}$.

  2. In (1), choose $(H,K)=(\langle r \rangle,\langle s \rangle)$. Then we have $cardinality(\langle r \rangle \langle s \rangle) = \frac{2n}{order(\langle r \rangle \cap \langle s \rangle)}$.

  3. Long story but I was able to prove that $\langle r \rangle \cap \langle s \rangle = 1_{D_n}$ (Outline of proof: If not, then contradiction if $n$ is odd. If not and if $n$ is even, then contradiction with $order(r)=n \ge 3$.)

  4. By (2) and (3), we have $cardinality(\langle r \rangle \langle s \rangle) = 2n$.

  5. Question 2: Ummm...does trivial intersection of subgroups imply the subgroups' product set is a subgroup?

If not, then I'm not sure this gets me anywhere. My idea is that for Part 1 we should have defined $D_n$ as of order $2n$ and then show that it has the form $$D_n = \{1_{D_n},r,r^2, ..., r^{n-1},$$ $$s, rs, r^2s, ..., r^{n-1}s\},$$

I would do this by:

  1. using the rule that for finite $[G:H], [G:K]$ even if $H$ and $K$ are no longer finite (they are finite still here though) we have $[G:H \cap K] = [G:H] [G:K]$ if and only if $G=HK$.

  2. and then concluding that $D_n = \langle r \rangle \langle s \rangle$, even though I think we weren't sure as to whether or not the product set $\langle r \rangle \langle s \rangle$ was a group (until showing equality with $D_n$).

BCLC
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    Are you familiar with group presentations and combinatorial group theory? – Shaun Oct 21 '21 at 12:38
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    A definition of $D_n$ is $D_{n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$ From this you can show that $D_n$ has $2n$ elements. – Dietrich Burde Oct 21 '21 at 12:43
  • @Shaun Yes, sort of, but it's not allowed for this class. – BCLC Oct 21 '21 at 12:43
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    Then it is also not allowed to assume a different definition than the one in your class. – Dietrich Burde Oct 21 '21 at 12:44
  • @DietrichBurde The class notes are little ambiguous as to what's defined or what's proven. It's not technically on any homework. It's just in the notes. Before I ask my instructor, I wanted to ask on stackexchange first. – BCLC Oct 21 '21 at 12:46
  • @DietrichBurde thanks! ok $2n$ is indeed not part of definition... 1 - so order(r)=n and order(s)=2 are even technically not part of definition of $D_n$. am i correct? 2 - so to prove $2n$ do i get anywhere above assuming I know the orders of $r$ and $s$ ? (I guess I'll have to refer to the keith conrad pdf's on dihedral if I don't assume orders hehe) – BCLC Oct 21 '21 at 12:47

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