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

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.

## closed as off-topic by Derek Holt, Christoph, Shaun, Alexander Gruber♦May 20 at 18:14

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$$D_4$$ 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 $$S_8$$.
• Also $D_8$ is a group of order $8$, so it is isomorphic to a subgroup of $S_8$. – bof May 20 at 6:53