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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

Allie
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    Good observation about the elements of order 12. What about the number of elements of other orders? – Calvin Lin Jun 10 '21 at 18:05
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    @CalvinLin so there are 6 elements of order 4 in $S_{4}$ and only 2 elements of order 4 in $D_{12}$ – Allie Jun 10 '21 at 18:10

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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

janmarqz
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jacopoburelli
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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

Dietrich Burde
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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.

Shaun
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