Show that a dihedral group of order $4$ is isomorphic to $V$, the $4$ group. 
Show that a dihedral group of order $4$ is isomorphic to $V$, the $4$ group.

Also, please show that a dihedral group of order $6$ is isomorphic to $S_3$.
Thank you!
 A: The dihedral group of order $2n$ is given by
$$
\begin{align}
  D_{2n} &= \langle r, s \mid ord(r) = n, ord(s) = 2, srs = r^{-1} \rangle \\
        &= \langle r, s \mid r^n = s^2 = (sr)^2 = 1 \rangle
\end{align}
$$
When $n=2$, we get
$$
D_{4} = \langle r, s \mid r^2 = s^2 = (sr)^2 = 1 \rangle
$$
This means that $D_4 = \{ 1, r, s, sr \}$. Since $r^2 = s^2 = (sr)^2 = 1$, $D_4$ is not cyclic and so must be $V$. 
When $n=3$, we get
$$
 D_{6} = \langle r, s \mid ord(r) = 3, ord(s) = 2, srs = r^{-1} \rangle 
$$
It is clear that $r \mapsto (123)$ and $s \mapsto (12)$ gives an isomorphism $D_6 \to S_3$.
A: the dihedral group of order $2n$ is generated by an element of order $n$ (call this $a$) and an involution (element of order 2) $r$ satisfying the relation $rar=a^{-1}$.
when the group has order 4 then $n=2$ hence $a^{-1} = a$. in this special case, therefore $r$ and $a$ commute since $rar=a$ implies, after right multiplication by $r$, that $rar^2 =ar$. finally $(ar)^2=arar=a^2r^2=1$.thus the three elements $r,a,ar$ commute, and each has order 2, so you should be able to convince yourself that the set $\{1,a,r,ar\}$ forms a group and satisfies the relations defining the viergruppe.
since this group is abelian it has no non-trivial inner automorphism, but you may easily show that its outer automorphism group is $S_3$, ie. the symmetric group of permutations of the three non-identity elements. show that this $S_3$ is generated by a transposition $\tau$ and a 3-cycle $\sigma$. then show that these two elements satisfy the relations $\tau^2=1, \sigma^3=1,\tau \sigma \tau = \sigma^{-1}$, and that these two elements therefore define the dihedral group of order $6$
