Is the symmetry group of the square isomorphic to $\mathbb{Z_8}$? to $\mathbb{S_8}$? to a subgroup of $\mathbb{S}_8$?

Question

Exercise: Mark True or False. Explain why.

a) The symmetry group of a square is isomorphic to $\mathbb{Z_8}$.

b) The symmetry group of a square is isomorphic to $\mathbb{S_8}$.

c)The symmetry group of a square is isomorphic to a subgroup $\mathbb{S_8}$.

I need explanation with isomorphism. I know that symmetry group of a square is dihedral group $\mathbb{D_4}$ which has 8 elements. $\mathbb{S_8}$ means all permutations. $\mathbb{Z_8}$ cyclic group of order 8.

Answer 1

$D_4$ is not abelian, but $\mathbb{Z}_8$ is. So they are not isomorphic.

$D_4$ has order $8$ but $S_8$ has order $8!$ and so they are not isomorphic.

$D_4$ is a permutation group of a set of cardinality $4$, so $D_4$ is isomorphic to a subgroup of $S_4$, and so a subgroup of $S_8$.