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? 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.
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.
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 $\mathbb Z_2 \times \mathbb Z_2$.