Prove $D_{12}$ and $S_4$ are not isomorphic groups in 3 different ways

Question

Give three reasons why $D_{12}$ and $S_{4}$ are not isomorphic. More precisely, prove that $D_{12}$ and every group isomorphic to it satisfy three properties that $S_{4}$ does not satisfy.

So I have proved that $D_{12}$ has an element of order $12$ and that $S_{4}$ does not. I am hoping to prove that there are a different number of elements of one order than in the other group but I don't know what else to do because neither are cyclic or abelian and the groups both have the same order.

Please help

Answer 1

I hope this helps : could be enough to note that $D_{12} \simeq \mathbb{Z}_{12} \rtimes_{\psi}\mathbb{Z}_2$ where $\psi$ is the inversion and $S_4 \simeq A_4 \rtimes_{\varphi} \mathbb{Z}_2$, where $\varphi$ is the conjugancy by a transposition

Answer 2

One property is that $S_4$ has trivial center, but $D_{12}$ has not.

References:

Prove that the symmetric group $S_n$, $n \geq 3$, has trivial center.

Center of dihedral group

Answer 3

You're correct that the element of order twelve in $D_{12}$ is one way to do this.

Two others include:

• $D_{12}$ has the derived subgroup isomorphic to $\Bbb Z_6$, whereas $S_4$ has derived subgroup isomorphic to $A_4$.

• $D_{12}$ has nine normal subgroups, whereas $S_4$ has four.