Is the group of invertible elements in $\mathbb{Z}/8\mathbb{Z}$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$ or to the symmetry group of the rectangle?

Question

Problem: Is the group $g(8) = \{[1], [3], [5], [7]\}$ isomorphic to $\mathbb{Z}_4$ or to the symmetry group of the rectangle?

Attempt: I know that $g(8)$ is isomorphic to $\mathbb{Z}_4$ because I compare the multiplication tables, and they look the same.

I know that $g(8)$ is not cyclic, and I know that the symmetry group of the rectangle is not cyclic since every non-identity element has order 2. Does that mean they are not isomorphic?

Can anyone help me verify the last part. Thank you.

Answer 1

We need to look at groups of $4$ elements: the set of invertible elements of $\mathbb Z_8$ is a multiplicative group of order four, and there are only two such groups of order four, up to isomorphism: $\mathbb Z_4,$ or the Klein-4 group.

So your group must be isomorphic to $\mathbb Z_4$ or else to the Klein-4 group.

Since your group is not cyclic, it cannot be isomorphic to $\mathbb Z_4$, which is cyclic. (Any group that is isomorphic to a cyclic group is necessarily also cyclic).

That leaves only the Klein-4 group, which is NOT cyclic: I'd suggest you compare the group tables of your group and the Klein-4 group to convince yourself that they are isomorphic. The Klein-4 group is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$.

Answer 2

There are only two groups of order four, up to isomorphism: $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. You can distinguish them because $\mathbb Z_4$ has two elements of order four, one element of order two and the identity. On the other hand, $\mathbb Z_2 \times \mathbb Z_2$ has three elements of order two and the identity.