$D_n$ is a group for all integers $n\geq 3$

Question

If one wants to prove that $D_n$ (a dihedral structure) is a group for all $n\geq 3$, would it be sufficient to state the following?

$D_n = \{1, s, r, ..., r^n, sr, ..., sr^n\}$.

1. Associativity: $(1s)r^i = sr^i=(sr^i)=1(sr^i)$

2. Identity: $1sr^i = sr^i1 = sr^i$.

3. Inverse: $s(r^i r^{-i})s = s1s = 1$.

I think that I'm not doing this proof completely properly, but I have no idea how else this can be proved. I would appreciate your evaluation of my proof. In particular, how does one prove that $D_2$ is not a group?

Answer 1

Ground Rules

Okay first it should be good to lay some ground rules about what we shouldn't have to prove. Which are the following (by definition of multiplication in $D_n$):

1. $sr=r^{-1}s$
2. $s^2=1=r^n$
3. If $d\in D_n$ then $d=s^ir^j$ for some $i,j\in Z$

with that as our ground rules.

How to prove $D_n$ is a group

[Quick note: $a_1,a_2,a_3,b_1,b_2,b_3$ are all just intergers]

1. (Associativty) Show that if $s^{a_1}r^{b_1},s^{a_2}r^{b_2},s^{a_3}r^{b_3}\in D_n$ then $s^{a_1}r^{b_1}(s^{a_2}r^{b_2}s^{a_3}r^{b_3})=(s^{a_1}r^{b_1}s^{a_2}r^{b_2})s^{a_3}r^{b_3}$

[To prove associativity first calculate whats inside the parathesis and do the next step example: (1+2)+3=(3)+3=6 and 1+(2+3)=1+(5)=6 thus (1+2)+3=1+(2+3) ]

2. (Closed) Show that if $s^{a_1}r^{b_1},s^{a_2}r^{b_2}\in D_n$ then $s^{a_1}r^{b_1}s^{a_2}r^{b_2}\in D_n$

[Prove that $s^{a_1}r^{b_1}s^{a_2}r^{b_2}$ can be rewritten as $s^{i}r^{j}$]

3. (identity) Show that there is an element $s^{a_1}r^{b_1}\in D_n$ such that for any $s^{a_2}r^{b_2}\in D_n$ we have that $s^{a_1}r^{b_1}s^{a_2}r^{b_2}=s^{a_2}r^{b_2}$

[Stems from the definition $1=s^{i}r^{j}$ where $i=2$ and $j=n$]

4. (inverses) Show that if $s^{a_1}r^{b_1}\in D_n$ there exist an element $s^{a_2}r^{b_2}\in D_n$ such that $s^{a_1}r^{b_1}s^{a_2}r^{b_2}=1$

[use definition 1 and 2 for this one ]

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Commentary and evaluation

First things first $D_2=\{1,r,s,sr\}$,$D_1=\{1,s\}$, $D_3=\{1,r,r^2,s,sr,sr^2\}$ and $D_0=\{1\}$ are all groups.

and secondly

Note that everything in $D_n$ is described as a power of $s$ times a power $r$.

1. So on your second part you can simply note that $1=s^{2}=r^n$ by definition (but your proof works fine)

2. your first part doesn't show that [$(r^i r^j)r^k=r^i(r^j r^k)$ and $(sr^i sr^j)sr^k=sr^i(sr^j sr^k)$].

3. You haven't proved closure either $sr^i sr^k\in D_n$ and $r^i r^k\in D_n$.