Show that $D_3 = S_3$ but $D_n \subsetneq S_n$ for $n \geq 4$.
In my class, we proved that $D_n$ is generated by $f = (1,2,3,...,n)$ and $g=(1)(2,n)(3,n-1)...(\frac{n+1}{2}, \frac{n+3}{2})$ (for an odd n) or $(1)(2,n)(3,n-1),...,(\frac{n}{2},\frac{n}{2}+2)(\frac{n}{2}+1)$ (for an even n) for $D_n = <f,g>$.
For the part where it says to show $D_3 = S_3$, I believe I can use the above information and get:
$$D_3 = <(1,2,3),(1)(2,3)>$$ = <(1,2,3),(1,2,3)>
I think this is equal to $(1,2,3)$ which is the same elements in $D_3$.
I don't know how to go about the second part, and would really like to know if the first part is even correct.
If we let $D_n$ act on the set of $n$ vertices of a regular $n$-gon, this gives us a homomorphism:
$D_n \to S_n$.
This action is faithful, as the only symmetry of the $n$-gon which fixes all the vertices is the identity map.
This shows $D_n \subseteq S_n$.
However, $|D_n| = 2n < n! = |S_n|$ if $n > 3$.