Dihedral and permutation groups isomorphism

Question

Why dihedral group $\mathbb{D_4}$ is isomorphic with $\mathbb{S_4}$ and not isomorphic with $\mathbb{S_8}$?

Answer 1

It's not isomorphic. $\Bbb D_4$ has $8$ elements (according to one convention; according to the other convention it has $4$ elements), while $\Bbb S_4$ has $24$ elemnents.

On the other hand, $\Bbb D_4$ is in a natural way isomorphic to a subgroup of $\Bbb S_4$, because they are both groups of symmetries on a set of $4$ elements.

Of course, a group of symmetries on $4$ elements can be seen as acting on $8$ elements, except it doesn't do anything to the last $4$ (or it does the same thing to the last $4$ as it does to the first $4$, in some variation), so both $\Bbb D_4$ and $\Bbb S_4$ are isomorphic to subgroups of $\Bbb S_8$ in this manner.

However, $\Bbb D_4$ is also isomorphic to a subgroup of $\Bbb S_8$ in a different way: There are $8$ elements in $\Bbb D_8$, and each element on $\Bbb D_8$ permutes the elements in $\Bbb D_8$ by multiplication (say from the left). In other words, $a\in \Bbb D_8$ is a permutation $\Bbb D_8\to\Bbb D_8$ defined by $a(b) = ab$. This makes it a subgroup of $\Bbb S_8$ in a different way to the way described above.

Answer 2

Unless you're using a non-standard definition of $\Bbb D_4,$ it isn't isomorphic to either of them. $\Bbb D_4$ has either $4$ or $2\cdot 4=8$ elements, depending on your definition, while $\Bbb S_4$ has $4!=24$ elements, and $\Bbb S_8$ has $8!=40320$ elements.

The dihedral group $\Bbb D_3$ happens to be isomorphic to $\Bbb S_3,$ but that's an exceptional case.