So I am trying to prove that $D_{12}$ and $S_4$ are not isomorphic by giving three properties that all isomorphisms to $D_{12}$ has that $S_4$ does not.
I have already proved that there is an element of order $12$ in $D_{12}$ and now I'm trying to prove that $D_{12}$ is not cyclic because I know $S_4$ is.
I know that there are at least two ways to do this: to find an element that does not divide the order of the group (there is a theorem stating that in a finite cyclic group the order of an element divides the order of the group) or to show that there is a subgroup of $D_{12}$ that is not cyclic (theorem stating every subgroup of a cyclic group is cyclic).
I can't find an element that does not divide the order $24$ and I'm honestly really confused on subgroups of Dihedral group.
Please help
Lemma: Each cyclic group is abelian.
Proof: Let $x,y\in G$ for a cyclic group $G$. Then $x=g^a, y=g^b$ for some $a,b\in\Bbb Z$, where $g$ is the generator of $G$. Now $xy=g^ag^b=g^{a+b}=g^{b+a}=g^bg^a=yx$.$\square$
Neither $D_{12}$ nor $S_4$ is abelian, so neither is cyclic.
Indeed, let $r$ be a nontrivial rotation in $D_{12}$ and $s$ be a flip. Then $srs=r^{-1}$, i.e., $sr=r^{-1}s$ but $r\neq r^{-1}$. Also,
$$\begin{align} (12)(13)&=(132)\\ &\neq (123)\\ &=(13)(12) \end{align}$$
in $S_4$.